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A SUITE OF CASE STUDIES IN RELATIONAL

DATABASE DESIGN

Master Thesis – Weiguang Zhang McMaster University- Computing and Software


A SUITE OF CASE STUDIES IN RELATIONAL DATABASE DESIGN

By

WEIGUANG ZHANG

Computer Science

A Thesis

Submitted to the School of Graduate Studies

In Partial Fulfillment of the Requirements

For the Degree

Master of Science

McMaster University

© Copyright by Weiguang Zhang January 2012


ii

MASTER OF SCIENCE (Jan, 2012) McMaster University

(Computer Science) Hamilton, Ontario

TITLE: A SUITE OF CASE STUDIES IN RELATIONAL DATABASE

DESIGN

AUTHOR: WEIGUANG ZHANG

SUPERVISOR: Dr. Antoine Deza and Dr. Frantisek Franek

NUMBER OF PAGES: x & 107


iii

Abstract

Typical relational database design examples in textbooks and undergraduate courses are

small and do not provide any real opportunity to practice the design, they simply illustrate

and illuminate the principles. On the other end of the spectrum are typical industrial

databases whose designs are complex and extensive, and so not suitable as a project for a

one term database course. The objective of this thesis is to design and develop a

collection of ten projects that would be usable as term projects in relational database

system design for a typical undergraduate database course. To this end a suite of ten case

studies are presented. Each project is taken from its informal specification to a relational

schema using entity-relationship modeling and its translation to relational model, to

database schema, to implementation of the database, to interactive SQL querying of the

installed database and finished with a simple application programmed in C using the

installed database and accessing it via embedded SQL.


iv

Acknowledgments

I would like to express my gratitude to all of those who made it possible to complete this

thesis, in particular to my supervisors Dr. Antoine Deza and Dr. Frantisek Franek.

I appreciate the great aid and support from all the members of the Advanced Optimization

Laboratory.

I would also like to thank my family for their understanding and continuous support.


v

Abbreviations

IBM: International Business Machines

ACM: Association of Computing Machinery

DBMS: Database Management System

SQL: Structured Query Language

ODBC: Open Database Connectivity

JDBC: Java Database Connectivity

ER: Entity Relationship

PHP: Personal Home Page

API: Application Programming Interface

CLI: Call Level Interface:

ESQL: Embedded Structured Query Language

IE: Information Engineering

IDEF1X: Integrated Definition for Information Modeling


vi

Terminologies and Symbols of ERwin IE Format

Identifying relationship: Shown with a solid line. An identifying relationship is a

relationship between two entities in which the child entity is dependent on its associated

parent entity, and the primary key of the parent entity is the part of the primary key of the

child entity.

Non-identifying Relationship: Shown with a dashed line. A non-identifying relationship

is a relationship between two entities in which the child entity is independent on its

associated parent entity, and the primary key of the parent entity is the non-key attribute

instead of the key attribute in the child entity.

Max relationship cardinality: Shown with a short perpendicular line across the

relationship near its line end to signify “one” and with a “crow’s foot” on the line end to

signify “many”.

Min relationship cardinality: Shown with a small circle near the end of the line to

signify “zero” (participation in the relationship is optional) or with a short perpendicular

line across the relationship line to signify “one” (participation in the relationship is

mandatory).

Dependent Entity:

Independent Entity:

Zero:

One:

Many:

Customer

Customer Email


vii

List of Figures

Figure 1: Summary of aspects of the Logical and Physical Models ............................... 7

Figure 2: The University System Logical Model ......................................................... 10

Figure 3: The University System Physical Model ........................................................ 11

Figure 4: The Airline Reservation Logical Model ....................................................... 27

Figure 5: The Airline Reservation Physical Model ...................................................... 28

Figure 6: The Movie Rental Logical Model ................................................................ 35

Figure 7: The Movie Rental Physical Model ............................................................... 36

Figure 8: The Car Rental Logic Model ........................................................................ 44

Figure 9: The Car Rental Physical Model .................................................................... 45

Figure 10: The Course Registration Logical Model ....................................................... 52

Figure 11: The Course Registration Physical Model...................................................... 53

Figure 12: The Emergency Room Logical Model.......................................................... 61

Figure 13: The Emergency Room Physical Model ........................................................ 62

Figure 14: The Property Rental Logical Model ............................................................. 70

Figure 15: The Property Rental Physical Model ............................................................ 71

Figure 16: The Software Project Logical Model ............................................................ 81

Figure 17: The Software Project Physical Model .......................................................... 82

Figure 18: The Tour Operator System Logical Model ................................................... 89

Figure 19: The Tour Operator System Physical Model .................................................. 90

Figure 20: The Warehouse System Logical Model ........................................................ 99

Figure 21: The Warehouse System Physical Model..................................................... 100


viii

Contents

Abstract ........................................................................................................................ iiii

Acknowledgments ....................................................................................................... ivv

Abbreviations .............................................................................................................. viv

Terminologies and Symbols of ERwin IE Format ...................................................... iv

List of Figures ............................................................................................................ viiii

Chapter 1: Introduction ................................................................................................ 1

Chapter 2: Objectives, Decisions, and Methods ........................................................... 4

Chapter 3: University System ....................................................................................... 8

3.1 UNIVERSITY SYSTEM INFORMAL DESCRIPTION .................................................................................8

3.2 UNIVERSITY SYSTEM LOGICAL MODEL .......................................................................................... 10

3.3 UNIVERSITY SYSTEM DB2 PHYSICAL MODEL ................................................................................ 11

3.4 UNIVERSITY SYSTEM DB2 SCHEMA ............................................................................................... 12

3.5 UNIVERSITY SYSTEM INTERACTIVE QUERIES ................................................................................. 14

3.6 UNIVERSITY SYSTEM APPLICATION PROGRAM ............................................................................... 16

Chapter 4: Airline Reservation ................................................................................... 25

4.1 AIRLINE RESERVATION INFORMAL DESCRIPTION ............................................................................ 25

4.2 AIRLINE RESERVATION LOGICAL MODEL ....................................................................................... 27

4.3 AIRLINE RESERVATION DB2 PHYSICAL MODEL ............................................................................. 28

4.4 AIRLINE RESERVATION DB2 SCHEMA ............................................................................................ 29

4.5 AIRLINE RESERVATION INTERACTIVE QUERIES ............................................................................... 31

Chapter 5: Movie Rental ............................................................................................. 34


ix

5.1 MOVIE RENTAL INFORMAL DESCRIPTION ....................................................................................... 34

5.2 MOVIE RENTAL LOGICAL MODEL .................................................................................................. 35

5.3 MOVIE RENTAL PHYSICAL DB2 MODEL......................................................................................... 36

5.4 MOVIE RENTAL DB2 SCHEMA ....................................................................................................... 37

5.5 MOVIE RENTAL INTERACTIVE QUERIES ......................................................................................... 39

Chapter 6: Car Rental ................................................................................................. 42

6.1 CAR RENTAL INFORMAL DESCRIPTION ........................................................................................... 42

6.2 CAR RENTAL LOGICAL MODEL ...................................................................................................... 44

6.3 CAR RENTAL PHYSICAL DB2 MODEL ............................................................................................ 45

6.4 CAR RENTAL DB2 SCHEMA ........................................................................................................... 46

6.5 CAR RENTAL INTERACTIVE QUERIES ............................................................................................. 48

Chapter 7: Course Registration .................................................................................. 51

7.1 COURSE REGISTRATION INFORMAL DESCRIPTION ........................................................................... 51

7.2 COURSE REGISTRATION LOGICAL MODEL ...................................................................................... 52

7.3 COURSE REGISTRATION PHYSICAL DB2 MODEL............................................................................. 53

7.4 COURSE REGISTRATION DB2 SCHEMA ........................................................................................... 54

7.5 COURSE REGISTRATION INTERACTIVE QUERIES.............................................................................. 58

Chapter 8: Emergency Room ...................................................................................... 60

8.1 EMERGENCY ROOM INFORMAL DESCRIPTION ................................................................................. 60

8.2 EMERGENCY ROOM LOGICAL MODEL ............................................................................................ 61

8.3 EMERGENCY ROOM PHYSICAL DB2 MODEL ................................................................................... 62

8.4 EMERGENCY ROOM DB2 SCHEMA ................................................................................................. 63

8.5 EMERGENCY ROOM INTERACTIVE QUERIES .................................................................................... 66

Chapter 9: Property Rental......................................................................................... 69

9.1 PROPERTY RENTAL INFORMAL DESCRIPTION ................................................................................. 69

9.2 PROPERTY RENTAL LOGICAL MODEL ............................................................................................. 70

9.3 PROPERTY RENTAL PHYSICAL DB2 MODEL ................................................................................... 71

9.4 PROPERTY RENTAL DB2 SCHEMA .................................................................................................. 72

9.5 PROPERTY RENTAL INTERACTIVE QUERIES .................................................................................... 77

Chapter 10: Software Project...................................................................................... 80

10.1 SOFTWARE PROJECT INFORMAL DESCRIPTION .............................................................................. 80

10.2 SOFTWARE PROJECT LOGICAL MODEL ......................................................................................... 81

10.3 SOFTWARE PROJECT PHYSICAL DB2 MODEL ................................................................................ 82

10.4 SOFTWARE PROJECT DB2 SCHEMA .............................................................................................. 83


x

10.5 SOFTWARE PROJECT INTERACTIVE QUERIES ................................................................................. 85

Chapter 11:Tour Operator System ............................................................................. 87

11.1 TOUR OPERATOR SYSTEM INFORMAL DESCRIPTION...................................................................... 87

11.2 TOUR OPERATOR SYSTEM LOGICAL MODEL ................................................................................. 89

11.3 TOUR OPERATOR SYSTEM PHYSICAL DB2 MODEL ................................................................................... 90

11.4 TOUR OPERATOR SYSTEM DB2 SCHEMA ...................................................................................... 91

11.5 TOUR OPERATOR SYSTEM INTERACTIVE QUERIES ........................................................................ 94

Chapter 12: Warehouse System .................................................................................. 97

12.1 WAREHOUSE SYSTEM INFORMAL DESCRIPTION ............................................................................. 97

12.2 WAREHOUSE SYSTEM LOGICAL MODEL ....................................................................................... 99

12.3 WAREHOUSE SYSTEM PHYSICAL DB2 MODEL............................................................................ 100

12.4 WAREHOUSE SYSTEM DB2 SCHEMA .......................................................................................... 101

12.5 WAREHOUSE SYSTEM INTERACTIVE QUERIES............................................................................. 103

Conclusion and Future Work.................................................................................... 106

Bibliography .............................................................................................................. 107


1

Chapter 1: Introduction

Database Management Systems are really ubiquitous in this age of Internet commerce.

Although the development of relational database system theory and practice can be traced

to the 1970 seminal paper A Relational Model of Data for Large Shared Data Banks by

E.F. Codd [1], the explosive growth of the use of relational database management systems

came with the advent of Internet. Databases not only represent significant infrastructure

for computer applications, but they also process the transactions and exchanges that drive

most of the world economy. A significant and growing segment of the software industry,

known as the database industry includes IBM Corporation, Oracle Corporation, Informix

Corporation, Sybase Incorporated, Teradata Corporation and Microsoft Corporation.

E. F. Codd found the database technologies of the late 1960’s taking the old-fashioned

view “that the burden of finding information should be on users ...” unsatisfactory. He

proposed what he called a “relational model” based on two fundamental premises: What

Codd called the "relational model" rested on two key points:

1. It provided means of describing data with its natural structure only—that is,

without any formatting aspects, i.e. superimposing any additional structure for

machine representation purposes.

2. Between application programs on the one hand and the database system on the

other. Accordingly, it provided a natural and consistent basis for a high level

query language which facilitated maximal independence

These aspects are utilized every second all over the world as most of the Internet

traffic and most of the Internet commerce rely on database access of some form. The

relational databases allow various applications in various programming languages to

access and modify a databases.

There are many drawbacks to the relational model and in many ways it is quite

obsolete, see [2] or in particular the ACM blog by M. Stonebraker, one of the pioneers of

the relational database management systems, [3]. In general , the disadvantages of the

relational approach can be summarized by:

1. They are slow and cumbersome; in the early days they were slow - relational

DBMS (database management systems) have to employ many tables to conform


2

to the various normalization rules. This can make them slow and resource

inefficient. However most contemporary relational DBMS do not have

performance problems now.

2. Restricted attribute sizes. Attribute lengths are usually capped by a maximum

size. This can lead to occasional practical problems e.g. a company with a 400

character name - not frequent, but it can happen.

3. SQL does not provide an efficient way to browse alphabetically through an index.

Thus some systems cannot provide a simple title A-Z browse.

4. To fine-tune a SQL query is not a simple task and its performance may often

depend on the SQL query engine having up-to-date statistics of the database.

5. They do not “storing” of objects, in essence, objects must be “disassembled”

before storing them in a relational database, and “assembled” when they are

fetched.

Yet, despite these deficiencies an obsoleteness, the use of relational databases

proliferates. The major advantages could be summarized as follows:

1. Overwhelmingly, the most popular type of DBMS in use and as a result technical

development effort ensures that advances appear quickly and reliably.

2. A multitude of third party tools that are designed to work with the popular

relational DBMS via standards such as Open Database Connectivity (ODBC) or

Java Database Connectivity (JDBC).

3. A multitude of third party design and modeling tools for the relational model, such

as ERwin© modeler used in this thesis.

4. Very well developed management tools and security with automatic data logging

and recovery.

5. Referential integrity controls ensuring data consistency.

6. Transactional processing ensuring that incomplete transactions do not occur.

Based on the robust demand for relational databases, all Canadian universities offer

database management course in their Computer Science programmes. A typical database

management course covers mostly the relational database systems and a significant

portion of the practical aspect of any such course deals with the modeling, design and


3

installation of relational databases, and querying of databases using SQL both

interactively and via application programs.

The majority of textbooks dealing with databases as well as the majority of on-line

database tutorials provide a comprehensive set of examples concerning all five aspects

mentioned above: modeling, design, and installation of a database, and interactive SQL

querying and SQL application programming. However, these examples are usually very

simple and very small, just to illustrate or illuminate a concept or principle. They are not

sufficient for practicing of these aspects. The real-world projects are on the other hand

rather complex and extensive, not suitable to a typical one-term undergraduate course in

database design and applications.

The objective of this thesis is to design and develop a suite of projects that are of

some substance to provide a real practice ground for modeling, design, and installation of

a relational database, and SQL interactive querying and SQL application programming,

yet be of a complexity and extent that would allow it to be used in a one-term

undergraduate course. To this end, a set of ten projects is presented, each leading to a

reasonably small database of 20 to 30 tables, yet exhibiting a comprehensiveness of larger

projects.

The structure of each project presented here is the same:

1. An informal specification of the project in simple English.

2. An Entity-Relationship (ER) model is constructed using Erwin © software

and presented in the form of what is called in Erwin’s terminology the Logical

and the Physical models.

3. The DB2 database schema is produced based on the ER model.

4. Several SQL queries that can be used to query of the installed database are

presented.

5. A simple application program in C is presented that utilize embedded SQL to

work with the installed database -- usually performing the same queries used in

the part 4.

Since the size of the thesis is already large enough and so only for the first project the

source code of the C application programs is presented, for all other projects the source

codes are listed in the Appendix.


4

Chapter 2: Objectives, Decisions,

and Methods

One of important decisions of this thesis was whether the classical ER modeling diagrams

were to be used, i.e. the diagrams as proposed by Peter Chen in his pivotal 1976 paper

The Entity-Relationship Model: Toward a Unified View of Data, see [4]. The typical

arguments in favour include the facts that majority of database textbooks present the

original diagrams and that the original diagrams facilitate better understanding of the

concepts and methods. They are simple to draw, so ER models can easily be done by

hand as long as the overall size is not too big.

On the other hand, the original diagrams are bulky and unwieldy, and so even a

simple design requires many pages. Moreover, the ER models must be translated to

relational model before a database schema can be produced. The industrial and

commercial software modelers typically use approach that is closer to the relational

model than to the ER model, and for good reasons. Graphically, such models are much

more compact and more wieldy and though conceptually a bit more challenging than the

ER approach, they are still relatively easy to learn and master. The arguments in favour of

employing the more industrial approach won and the models in this thesis were created

using a professional design software Erwin ©

.

A decision concerning application programming affected the selection of the DBMS

for our projects. Any application program must access a database, usually using a

software tool or an interface provided by the vendor of the DBMS. Among languages

popular for database application programming, Python and PHP stand out. Python

accesses the database using API -- application programming interface. If Python was just

chosen as the language of application programming, the students would have to know or

learn Python and learn the appropriate API -- a too steep requirement to fit into a single

term, database course that should be focused on the DBMS rather than Python or API of

the DBMS. PHP access the database using Microsoft ODBC. Similarly as for Python, the

students would have to know or learn PHP and learn the “language” of ODBC, moreover

the ODBC is a software that must be purchased independently from the DBMS. Another

popular language for database application is Java that accesses a database using JDBC.

Even though this is a less stringent requirement than PHP and ODBC, it still is too

stringent.


5

Using embedded SQL in C programs is the least stringent requirement as the students

do not have to learn any new language as by the time they take the database course,

practically all students know enough of C, they cover SQL in the database course, so the

expenditure for learning both static embedded SQL and dynamic embedded SQL is

significantly less than learning ODBC, or JDBC, or API etc. Embedded SQL, though not

limited to C, works best with C as the CLI -- call level interface -- to which embedded

SQL are translated by the pre-processor is well defined in C as they are all actually

programmed in C. For these reasons, we opted for application programming based on

embedded SQL and C.

Another important decision concerned which of the concrete database management

systems (DBMS for short) should be used for thesis. The original idea to include several

major ones -- Oracle, DB2, and MySQL -- was rejected as the resulting thesis would be

just too large. For each system, we looked at several criteria:

• How available it is to students, and McMaster students in particular?

• How commercially successful and practical the system is?

• Does ERwin provide modeling support for it?

• Does the system provide a native support -- i.e. set of tools -- for embedded

SQL programming in C?

On availability to students, all three DBMS considered scored equally as they all offer

free versions: DB2 has a free student version, Oracle provided free Oracle Database

Express Edition, and MySQL that started as an Open Software is in public domain. On

particular availability to McMaster students DB2 scored the highest as it is the DBMS of

choice for the undergraduate database course and its graduate variant.

In commercial success and practicality, Oracle is a bit ahead of DB2 that is a bit ahead

of MySQL, however the differences were not significant enough for our purposes.

Contemporary version of ERwin provides modeling support for all three DBMS

considered equally.

For embedded SQL (ESQL for short), both DB2 and Oracle provide a very good

support, while MySQL lacks any native support for ESQL, third-party providers have

been promising ESQL tools (pre-processor) for some time, but none is still available.

Based on the above criteria and the fact that due to the size, for this thesis we had to

consider a single DBMS, DB2 was chosen over Oracle due to better availability in

particular to McMaster students. MySQL was rejected for the lack of ESQL support.


6

For the data modeling part of our projects, we decided to use ERwin modeling tool. In

real world, most data models are the form of ER diagrams that visually represent entities,

attributes, and relationships. As mentioned in the introduction, the original Chen’s format

is not very suitable as it is too bulky. Most of available software modeling tools thus

provide ER models that are closer to the relational model than to the pure ER model.

ERwin is no exception, and supports both the Information Engineering format (IE, most

popular format) and the IDEF1X format. Moreover ERwin is user friendly and easy to

learn, provides free Community Edition of Erwin Data Modeler, and for most of the

DBMS we considered generates database schemas automatically.

ERwin models have two forms, so-called Logical Model and Physical Model. The

Logical Model is a representation of an organization’s data, organized in terms of entities

and relationships and independent of any particular data management technology. It

includes all entities and relationships among them, all primary keys for all entities are

specified, i.e. weak entities are resolved, all attributes for each entity are specified, all

foreign keys are specified. Normalization occurs at this level, if needed.

The basic steps in producing the Logical Model are:

• Specify all entities.

• Find attributes for each entity.

• Specify primary keys for all entities.

• Find the relationships between different entities.

• Resolve many-to-many relationships.

• Perform normalization.

The Physical Model represents how the model will be built/stored in the database. A

physical database model shows all table structures, including column names, column data

types, column constraints, primary keys, foreign keys, and relationships between tables .

The features of the Physical Model include the specification all tables and columns,

foreign keys are used to identify relationships between tables, intentional de-

normalization may occur at this level based on user requirements. The physical

considerations may cause the Physical Model to be quite different from the Logical

Model, thus the Physical Model will be different for different DBMS. For example, data

type for a column maybe different between DB2 and SQL Server.

The basic steps in producing the Physical Model are:

• Convert entities into tables.

• Convert relationships into foreign keys.


7

• Convert attributes into columns.

• Modify the physical data model based on physical constraints/requirements.

Figure 1: Summary of aspects of the Logical and Physical Models


8

Chapter 3: University System

3.1 University System Informal Description

ABC University is a large institution with several campuses. Each campus has a different

name, address, distance to the city center and the only bus running to the campus. Each

campus has one club. The name of the club, the building in which the club is located, the

phone number of the club and the multiple sports which club offers, should all be

recorded.

The University consists of a number of faculties, such as the Art Faculty, the

Science Faculty, and so on. Each faculty has a name, dean and building. A faculty may be

divided into a number of schools, for example, the Science Faculty has a School of

Physics and a School of Chemistry. Each school belongs to one faculty only and is

located on just one campus, but one campus maybe the location of many schools.

Every school has name and an building assigned to. Each school offers different

programmes and each programme can be offered by only one school. Each programme

has a unique code, title, level and duration. Each programme comprises several courses,

different programmes have different courses. Each course has a unique code and course

title. Some courses may have one or more prerequisite courses and one course can be the

prerequisite course of some other courses.

Each of the students is enrolled in a single programme of study which involves a

fixed core of courses specific to that programme as well as a number of electives taken

from other programmes. Students work on courses and are awarded a grade in any course

if he/she passes the course. Otherwise the student has to re-take the failed course. The

system needs to record the year and term in which the course was taken and the grade

awarded to the student. Every student has a unique ID. The system also keeps the student

name, birthday and the year he/she enrolled in the course.

The school employs lecturers to teach the students. A lecturer is allowed to work

for one school only. Each lecturer is assigned an ID which is unique across the whole

university. The system keeps the lecturer’s name, title and the office room. A supervisor

maybe in charge of several lecturers, but a lecturer, however reports to only one

supervisor. A lecturer can teach many different courses. A course may also have been

taught by many different lecturers.


9

The university is operated by committees. Each faculty has to have a number of

committees with the same titles across the university, such as the Faculty Executive, the

Post Graduate Studies Committee, the Health and Sanity Committee, and so on. The

committees meet regularly, such as weekly or monthly. The frequency is determined by

the faculty involved. A committee’s members are all lecturers. A lecturer may be a

member of several committees.


10

3.2 University System Logical Model

Figure 2: The University System Logical Model


11

3.3 University System DB2 Physical Model

Figure 3: The University System Physical Model


12

3.4 University System DB2 Schema

CREATE TABLE CAMPUS(

CMPSNM VARCHAR(20) NOT NULL,

CMPSADDR VARCHAR(50),

DIST NUMERIC(8, 2),

BUSNO VARCHAR(10),

PRIMARY KEY (CMPSNM)

)

CREATE TABLE FACULTY(

FACNM VARCHAR(30) NOT NULL,

DEANNM VARCHAR(20),

FACBLD VARCHAR(20),

PRIMARY KEY (FACNM)

)

CREATE TABLE SCHOOL(

SCHLNM VARCHAR(30) NOT NULL,



SCHLBLD VARCHAR(20),

PRIMARY KEY (SCHLNM),

FOREIGN KEY (CMPSNM) REFERENCES CAMPUS (CMPSNM),

FOREIGN KEY (FACNM) REFERENCES FACULTY (FACNM)

)

CREATE TABLE PROGRAMME(

PROGCD CHAR(11) NOT NULL,


PROGTITL VARCHAR(20),

PROGLVL VARCHAR(10),

PROGLEN VARCHAR(20),

PRIMARY KEY (PROGCD),

FOREIGN KEY (SCHLNM) REFERENCES SCHOOL (SCHLNM)

)

CREATE TABLE STUDENT(

STUID INTEGER NOT NULL,


STUNM VARCHAR(30),

STUBRTH DATE,

YRENRL INTEGER,

PRIMARY KEY (STUID),

FOREIGN KEY (PROGCD) REFERENCES PROGRAMME (PROGCD)

)

CREATE TABLE COURSE(

CRSECD CHAR(10) NOT NULL,


CRSETITL VARCHAR(20),

PRIMARY KEY (CRSECD),

FOREIGN KEY (PROGCD) REFERENCES PROGRAMME (PROGCD)

)


13

CREATE TABLE COURSE_STUDENT(


STUID INTEGER NOT NULL,

YRTKN CHAR(4),

SEMTKN VARCHAR(20),

GRDAWRD VARCHAR(10),

PRIMARY KEY (CRSECD, STUID),

FOREIGN KEY (STUID) REFERENCES STUDENT (STUID),

FOREIGN KEY (CRSECD) REFERENCES COURSE (CRSECD)

)

CREATE TABLE CLUB(

CLBNM VARCHAR(20) NOT NULL,


CLBBLD VARCHAR(20),

PHNNO CHAR(12),

PRIMARY KEY (CLBNM),

FOREIGN KEY (CMPSNM) REFERENCES CAMPUS (CMPSNM)

)

CREATE TABLE SPORT(

SPRTNM VARCHAR(20) NOT NULL,

CLBNM VARCHAR(20) NOT NULL,

PRIMARY KEY (SPRTNM, CLBNM),

FOREIGN KEY (CLBNM) REFERENCES CLUB (CLBNM)

)

CREATE TABLE PRE_COURSE(


PRECRSECD CHAR(10) NOT NULL,

PRIMARY KEY (CRSECD, PRECRSECD),

FOREIGN KEY (CRSECD) REFERENCES COURSE (CRSECD),

FOREIGN KEY (PRECRSECD) REFERENCES COURSE (CRSECD)

)

CREATE TABLE LECTURER(

STFID INTEGER NOT NULL,


SUPID INTEGER,

LECTNM VARCHAR(20),

LECTTITL VARCHAR(30),

OFFROOM VARCHAR(10),

PRIMARY KEY (STFID),

FOREIGN KEY (SCHLNM) REFERENCES SCHOOL (SCHLNM),

FOREIGN KEY (SUPID) REFERENCES LECTURER (STFID)

)

CREATE TABLE LECTURER_COURSE(



PRIMARY KEY (STFID, CRSECD),

FOREIGN KEY (CRSECD) REFERENCES COURSE (CRSECD),

FOREIGN KEY (STFID) REFERENCES LECTURER (STFID)

)

CREATE TABLE COMMITTEE(

COMMTITL VARCHAR(30) NOT NULL,


MTFREQ VARCHAR(10),

PRIMARY KEY (COMMTITL, FACNM),


14

FOREIGN KEY (FACNM) REFERENCES FACULTY (FACNM)

)

CREATE TABLE COMMITTEE_LECTURER(


COMMTITL VARCHAR(30) NOT NULL,


PRIMARY KEY (STFID, COMMTITL, FACNM),

FOREIGN KEY (STFID) REFERENCES LECTURER (STFID),

FOREIGN KEY (COMMTITL, FACNM) REFERENCES COMMITTEE (COMMTITL,

FACNM)

)

3.5 University System Interactive Queries

Query 1: List all the schools are located in 'Toronto Campus', and sort them by school

name.

SELECT SCHLNM AS SCHOOL_NAME

FROM SCHOOL

WHERE CMPSNM = 'Toronto Campus'

ORDER BY SCHLNM

Query 2: List all the programmes provided by 'science faculty'.

SELECT P.PROGCD AS PROGRAMME_CODE

FROM PROGRAMME P

INNER JOIN SCHOOL S ON P.SCHLNM = S.SCHLNM

INNER JOIN FACULTY F ON S.FACNM = F.FACNM

WHERE F.FACNM = 'science faculty'

Query 3: Give all the names of the lecturers who are the members of the committee and

sort by their name.

SELECT DISTINCT L.LECTNM AS LECTURER_NAME

FROM COMMITTEE_LECTURER CL

JOIN LECTURER L ON CL.STFID = L.STFID

ORDER BY L.LECTNM

Query 4: List all supervisor's name and the name of the lecturer they manage. Please sort

by supervisor name and lecturer name.

SELECT SUP.LECTNM AS SUPERVISOR_NAME,L.LECTNM AS LECTURER_NAME

FROM LECTURER SUP

INNER JOIN LECTURER L ON SUP.STFID = L.SUPID

ORDER BY SUP.LECTNM,L.LECTNM

Query 5: Give all the lecturers who are not the member of the committee.


15

SELECT STFID AS STAFF_ID

FROM LECTURER

WHERE STFID NOT IN (SELECT DISTINCT STFID FROM COMMITTEE_LECTURER)

Query 6: Give the total number of courses for each programme.

SELECT PROGCD AS PROGRAMME_CODE,COUNT(CRSECD) AS NUMBER_OF_COURSE

FROM COURSE

GROUP BY PROGCD

Query 7: Give all the lecturers with the courses they are teaching. Sort by lecturer name.

SELECT L.LECTNM AS LECTURER_NAME,C.CRSETITL AS COURSE_TITLE

FROM LECTURER_COURSE LC

INNER JOIN LECTURER L ON LC.STFID = L.STFID

INNER JOIN COURSE C ON LC.CRSECD = C.CRSECD

ORDER BY L.LECTNM

Query 8: Give all the course titles and their corresponding prerequisite course titles.

SELECT C1.CRSETITL AS COURSE_TITLE,C2.CRSETITL AS PRE_COURSE_TITLE

FROM PRE_COURSE PC

INNER JOIN COURSE C1 ON PC.CRSECD = C1.CRSECD

INNER JOIN COURSE C2 ON PC.PRECRSECD = C2.CRSECD

Query 9: Give the top 5 courses which have more students involved.

SELECT C.CRSECD AS COURSE_CODE,COUNT(SS.STUID) AS NUMBER_OF_STUDENTS

FROM COURSE_STUDENT SS

LEFT JOIN COURSE S ON SS.SUBJCD = S.SUBJCD

LEFT JOIN COURSE C ON S.CRSECD = C.CRSECD

GROUP BY C.CRSECD

ORDER BY NUMBER_OF_STUDENTS DESC

FETCH FIRST 5 ROWS ONLY

Query 10: Give any of the prerequisite courses was not took by any of the students who

enrolled into the university in 2010, and were taking the courses in 2011.

SELECT DISTINCT STUID,PC.PRECRSECD AS PRE_COURSE

FROM COURSE_STUDENT CS

RIGHT JOIN PRE_COURSE PC ON CS.CRSECD = PC.CRSECD

WHERE STUID IN (SELECT STUID FROM STUDENT WHERE YRENRL = 2010)

AND YRTKN = 2011

EXCEPT

SELECT STUID,CRSECD

FROM COURSE_STUDENT


AND YRTKN = 2010


16

3.6 University System Application Program

Write a simple C program using embedded SQL. The program is to excute and display

the results on the screen in the same format as the DB2. It is to execute any of the ten

queries listed in the previous section based on what action the user chooses, or all ten

queries together. Assume the name of the databse is OURDB.

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

EXEC SQL INCLUDE SQLCA;

void query1();

void query2();

void query3();

void query4();

void query5();

void query6();

void query7();

void query8();

void query9();

void query10();

void do_all();

int connected=0;

void sqlerr(char* x) {

if (SQLCODE!=0) {

printf("error %s (%d)\n",x,SQLCODE);

if (connected) {

EXEC SQL CONNECT RESET;

printf("disconnected from OURDB\n");

}

exit(1);

}

}

void connect() {

EXEC SQL CONNECT TO OURDB;

sqlerr("CONNECT TO OURDB");

printf("Connected to OURDB\n");

connected = 1;

}

void disconnect() {

EXEC SQL CONNECT RESET;

sqlerr("CONNECT RESET");

printf("Disconnected from CS3DB3\n");

connected = 0;

}


17

void depad(char* s) {

int i;

i=strlen(s)-1;

while(s[i]==' ') s[i--]='\0';

}

char buffer[300];

int main() {

int i, j;

connect();

while(1) {

A: printf("enter number of query you want to execute "

"(1-10),\n 0 to quit, anything else all queries\n");

fgets(buffer,300,stdin);

if (buffer[0]=='\n') {

printf("nothing entered\n");

continue;

}

i=j=0;

while(buffer[j]!='\n') {

if (buffer[j]>='0' && buffer[j]<='9')

i = 10*i+buffer[j]-'0';

else{

printf("incorrect entry\n");

goto A;

}

j++;

}

if (i == 0) break;

else if (i == 1) query1();










else {

do_all();

break;

}

}

disconnect();

return 0;

}

int count, len, len1;

char buf1[80];

EXEC SQL BEGIN DECLARE SECTION;


18

char school_name[30];

char programme_code[11];

char lecturer_name[20];

char supervisor_name[20];

sqlint32 staff_id;

sqlint16 number_of_course;

char course_title[20];

char pre_course_title[20];

sqlint16 number_of_students;

sqlint32 student_id;

char pre_course[10];

EXEC SQL END DECLARE SECTION;

void query1() {

printf("displaying result of query 1\n");

EXEC SQL DECLARE cur1 CURSOR FOR

SELECT SCHLNM AS SCHOOL_NAME

FROM SCHOOL

WHERE CMPSNM = 'Toronto Campus'

ORDER BY SCHLNM;

sqlerr("OPEN cur1");

count=0;

while(1) {

EXEC SQL FETCH cur1 INTO :school_name;

if (SQLCODE!=0) break;

if (count==0) {

printf("SCHOOL_NAME\n");

printf("-----------------\n");

}

depad(school_name);

printf("%s\n",school_name);

count++;

}

EXEC SQL CLOSE cur1;

sqlerr("CLOSE cur1");

printf("\n %d record(s) selected.\n",count);

}

void query2() {



SELECT P.PROGCD AS PROGRAMME_CODE

FROM PROGRAMME P

INNER JOIN SCHOOL S ON P.SCHLNM = S.SCHLNM

INNER JOIN FACULTY F ON S.FACNM = F.FACNM

WHERE F.FACNM = 'science faculty';

sqlerr("DECLARE cur2");

EXEC SQL OPEN cur2;



19

count=0;

while(1) {

EXEC SQL FETCH cur2 INTO :programme_code;


if (count==0) {

printf("PROGRAMME_CODE\n");

printf("---------------\n");

}

depad(programme_code);

printf("%s\n",programme_code);

count++;

}




}

void query3() {



SELECT DISTINCT L.LECTNM AS LECTURER_NAME

FROM COMMITTEE_LECTURER CL

JOIN LECTURER L ON CL.STFID = L.STFID

ORDER BY L.LECTNM;


EXEC SQL OPEN cur3;


count=0;

while(1) {

EXEC SQL FETCH cur3 INTO :lecturer_name;


if (count==0) {

printf("LECTURER_NAME\n");

printf("------------------\n");

}

printf("%s\n",lecturer_nam);

count++;

}




}

void query4() {

int i;




20

SELECT SUP.LECTNM AS SUPERVISOR_NAME,L.LECTNM AS LECTURER_NAME

FROM LECTURER SUP

INNER JOIN LECTURER L ON SUP.STFID = L.SUPID

ORDER BY SUP.LECTNM,L.LECTNM;


EXEC SQL OPEN cur4;


count=0;

while(1) {

EXEC SQL FETCH cur4 INTO :supervisor_name,

:lecturer_name;


if (count==0) {

printf("SUPERVISOR_NAME LECTURER_NAME\n");

printf("------------------------------------------\n");

}

depad(supervisor_name);

depad(lecturer_name);

printf("%s",supervisor_name);

for(i = 20-strlen(supervisor_name); i>=0; i--) fputc(' ',stdout);

printf("%s\n",lecturer_name);

count++;

}




}

void query5() {



SELECT STFID AS STAFF_ID

FROM LECTURER

WHERE STFID NOT IN

(SELECT DISTINCT STFID FROM COMMITTEE_LECTURER);


EXEC SQL OPEN cur5;


count=0;

while(1) {

EXEC SQL FETCH cur5 INTO :staff_id;


if (count==0) {

printf("STAFF_ID\n");

printf("-----------\n");

}

depad(staff_id);

printf("%s\n",staff_id);

count++;


21

}




}

void query6() {

int i;



SELECT PROGCD AS PROGRAMME_CODE,COUNT(CRSECD) AS NUMBER_OF_COURSE

FROM COURSE

GROUP BY PROGCD;


EXEC SQL OPEN cur6;


count=0;

while(1) {

EXEC SQL FETCH cur6 INTO :programme_code,

:number_of_course;


if (count==0) {

printf("PROGRAMME_CODE NUMBER_OF_COURSE\n");

printf("-----------------------------------------\n");

}


depad(number_of_course);

printf("%s",programme_code);

for(i = strlen(programme_code); i<11; putchar(' '), i++);

printf("%s",number_of_course);

count++;

}




}

void query7() {

int i;



SELECT L.LECTNM AS LECTURER_NAME,C.CRSETITL AS COURSE_TITLE

FROM LECTURER_COURSE LC

INNER JOIN LECTURER L ON LC.STFID = L.STFID

INNER JOIN COURSE C ON LC.CRSECD = C.CRSECD

ORDER BY L.LECTNM;



22

EXEC SQL OPEN cur7;


count=0;

while(1) {

EXEC SQL FETCH cur7 INTO :lecturer_name,

:course_title;


if (count==0) {

printf("LECTURER_NAME COURSE_TITLE\n");

printf("-------------------------------------\n");

}

depad(lecturer_name);

depad(course_title);

printf("%s",lecturer_name);

for(i = strlen(lecturer_name); i<9; putchar(' '), i++);

printf("%s",course_title);

count++;

}




}

void query8() {

int i;



SELECT C1.CRSETITL AS COURSE_TITLE,C2.CRSETITL AS

PRE_COURSE_TITLE

FROM PRE_COURSE PC

INNER JOIN COURSE C1 ON PC.CRSECD = C1.CRSECD

INNER JOIN COURSE C2 ON PC.PRECRSECD = C2.CRSECD;


EXEC SQL OPEN cur8;


count=0;

while(1) {

EXEC SQL FETCH cur8 INTO :course_title,

:pre_course_title;


if (count==0) {

printf("COURSE_TITLE PRE_COURSE_TITLE\n");

printf("--------- ---------------------------------\n");

}

depad(course_title);

printf("%s",course_title);

for(i = strlen(course_title); i<10; putchar(' '), i++);

printf("%.24f\n",pre_course_title);


23

count++;

}




}

void query9() {

int i;



SELECT P.PROGCD AS PROGRAMME_CODE,COUNT(CS.STUID) AS

NUMBER_OF_STUDENTS


LEFT JOIN COURSE C ON CS.CRSECD = C.CRSECD

LEFT JOIN PROGRAMME P ON P.PROGCD = C.PROGCD

GROUP BY P.PROGCD

ORDER BY NUMBER_OF_STUDENTS DESC

FETCH FIRST 5 ROWS ONLY;


EXEC SQL OPEN cur9;


count=0;

while(1) {

EXEC SQL FETCH cur9 INTO :programme_code,

:number_of_students;


if (count==0) {

printf("PROGRAMME_CODE NUMBER_OF_STUDENTS\n");

printf("----------------------------------------\n");

}


depad(number_of_students);

printf("%s",programme_code);

for(i = strlen(programme_code); i<12; putchar(' '), i++);

printf("%s\n",number_of_students);

count++;

}




}

void query10() {

int i;



24


SELECT DISTINCT STUID,PC.PRECRSECD AS PRE_COURSE


RIGHT JOIN PRE_COURSE PC ON CS.CRSECD = PC.CRSECD


AND YRTKN = 2011

EXCEPT

SELECT STUID,CRSECD

FROM COURSE_STUDENT


AND YRTKN = 2010;


EXEC SQL OPEN cur10;


count=0;

while(1) {

EXEC SQL FETCH cur10 INTO :student_id,

:pre_course;


if (count==0) {

printf("STUDENT_ID PRE_COURSE\n");

printf("-----------------------------------\n");

}

depad(student_id);

printf("%s",student_id);

for(i = strlen(student_id); i<10; putchar(' '), i++);

printf("%d\n",pre_course);

count++;

}




}

void do_all() {

query1();

query2();

query3();

query4();

query5();

query6();

query7();

query8();

query9();

query10();

}


25

Chapter 4: Airline Reservation

4.1 Airline Reservation informal description

There are 6 different airlines in 6 different countries: Canada – AirCan, USA - USAir,

UK - BritAir, France - AirFrance, Germany - LuftAir, Italy - ItalAir. Their flights involve

the following 12 cities: Toronto and Montreal in Canada, New York and Chicago in US,

London and Edinburgh in UK, Paris and Nice in France, Bonn and Berlin in Germany,

Rome and Naples in Italy. In each of the 12 cities, there is a (single) booking office. You

are going to design a central air-reservation database to be used by all booking offices.

The flight has a unique flight number, air line code, business class indicator,

smoking allowed indicator. Flight availability has flight number, date + time of departure,

number of total seats available in business class, number of booked seats in business

class, number of total seats available in economy class, and number of booked seats in

economy class.

The customers may come from any country, not just the 6 above, and from any

province/state, and from any city. Customer has first & last name, mailing address, zero

or more phone numbers, zero or more fax numbers, and zero or more email addresses.

Mailing address has street, city, province or state, postal code and country. Phone/fax

number has country code, area code and local number. Email address has only one string,

and no structure is assumed. A customer can book one or more flights. Two or more

customers may have same mailing address and/or same phone number(s) and/or same fax

number(s). But the email address is unique for each customer. First and last names do not

have to be unique.

Booking has an unique booking number, booking city, booking date, flight

number, date + time of departure (in local time, and time is always in hours and minutes),

date + time of arrival (in local time), class indicator, total price (airport tax in origin +

airport tax in destination + flight price – in local currency. The flight price for business

class is 1.5 times of the listed flight price), status indicator (three types: booked. Canceled

– the customer canceled the booking, scratched – the customer had not paid in full 30

days prior to the departure), customer who is responsible for payment, amount-paid-so far

(in local currency), outstanding balance (in local currency), the first & last names to be

printed on the ticket. The airport taxes must be stored in local currencies (i.e. Canadian


26

dollars, US dollars, British Pounds, French francs, German marks, and Italian Liras).

Since the exchange rates change daily, they also must be stored for calculations of all

prices involved.

Though France, Germany, and Italy have had a common currency for a while, we

used the names of their original currencies to involve in this exercise currency exchange

rates and their changes.


27

4.2 Airline Reservation Logical Model

Figure 4: The Airline Reservation Logical Model


28

4.3 Airline Reservation DB2 Physical Model

Figure 5: The Airline Reservation Physical Model


29

4.4 Airline Reservation DB2 Schema

CREATE TABLE CUSTOMER (

CUSTID INT NOT NULL,

FNAME VARCHAR(20) NOT NULL,

LNAME VARCHAR(20) NOT NULL,

STREET VARCHAR(50) NOT NULL,

CITY VARCHAR(30) NOT NULL,

PROVINCE VARCHAR(30) NOT NULL,

COUNTRY VARCHAR(30) NOT NULL,

POSTCODE VARCHAR(20) NOT NULL,

PRIMARY KEY (CUSTID)

)

CREATE TABLE PHONE (

PCRTYCODE CHAR(2) NOT NULL,

PAREACODE CHAR(3) NOT NULL,

PNUMBER CHAR(7) NOT NULL,


PRIMARY KEY (CUSTID, PCRTYCODE, PAREACODE, PNUMBER),

FOREIGN KEY (CUSTID) REFERENCES CUSTOMER (CUSTID)

)

CREATE TABLE FAX (

FAREACODE CHAR(3) NOT NULL,

FCTRYCODE CHAR(2) NOT NULL,

FNUMBER CHAR(7) NOT NULL,


PRIMARY KEY (CUSTID, FCTRYCODE, FAREACODE, FNUMBER),


)

CREATE TABLE EMAIL (

EMAIL VARCHAR(50) NOT NULL,


PRIMARY KEY (CUSTID, EMAIL),

UNIQUE (EMAIL),


)

CREATE TABLE COUNTRY (

CTRYCD CHAR(2) NOT NULL,

CTRYNM VARCHAR(30),

PRIMARY KEY (CTRYCD)

)

CREATE TABLE AIRLINE (

AIRLINECD VARCHAR(10) NOT NULL,


PRIMARY KEY (AIRLINECD),

FOREIGN KEY (CTRYCD) REFERENCES COUNTRY (CTRYCD)

)

CREATE TABLE FLIGHT (

FNO VARCHAR(10) NOT NULL,

SMALLOW CHAR(1) NOT NULL,

BCAVL CHAR(1),

AIRLINECD VARCHAR(10) NOT NULL,


30

PRIMARY KEY (FNO),

FOREIGN KEY (AIRLINECD) REFERENCES AIRLINE (AIRLINECD)

)

CREATE TABLE CURRENCY (

FCURR CHAR(3) NOT NULL,

TCURR CHAR(3) NOT NULL,

EXCHRATE DECIMAL(8, 4) NOT NULL,

PRIMARY KEY (FCURR, TCURR)

)

CREATE TABLE CITY (

CITYNM VARCHAR(30),

CITYID INT NOT NULL,


PRIMARY KEY (CITYID),

FOREIGN KEY (CTRYCD) REFERENCES COUNTRY (CTRYCD)

)

CREATE TABLE AIRPORT (


AIRPORTCD CHAR(3) NOT NULL,

AIRPORTNM VARCHAR(50),

AIRPORTTAX DECIMAL(6, 2) NOT NULL,

PRIMARY KEY (AIRPORTCD),

FOREIGN KEY (CITYID) REFERENCES CITY (CITYID)

)

CREATE TABLE CITY_CURRENCY (


FCURR CHAR(3) NOT NULL,

TCURR CHAR(3) NOT NULL,

PRIMARY KEY (CITYID,FCURR,TCURR),

FOREIGN KEY (CITYID) REFERENCES CITY (CITYID),

FOREIGN KEY (FCURR,TCURR) REFERENCES CURRENCY (FCURR,TCURR)

)

CREATE TABLE FLIGHT_AVAILABILITY (

FLEN VARCHAR(30) NOT NULL,

DEPTTIME VARCHAR(30) NOT NULL,

ARRTIME VARCHAR(30) NOT NULL,


BBUSSEAT INT NOT NULL,

BECOSEAT INT NOT NULL,

TBUSSEAT INT NOT NULL,

TECOSEAT INT NOT NULL,

DEST CHAR(3) NOT NULL,

ORIG CHAR(3) NOT NULL,

PRIMARY KEY (FNO, ORIG, DEST, DEPTTIME, ARRTIME),

FOREIGN KEY (FNO) REFERENCES FLIGHT (FNO),

FOREIGN KEY (DEST) REFERENCES AIRPORT (AIRPORTCD),

FOREIGN KEY (ORIG) REFERENCES AIRPORT (AIRPORTCD)

)

CREATE TABLE CLASS (

CLASSID INT NOT NULL,

CLASS VARCHAR(10),

PRIMARY KEY (CLASSID)

)

CREATE TABLE STATUS (


31

STATUSID INT NOT NULL,

STATUS VARCHAR(10),

PRIMARY KEY (STATUSID)

)

CREATE TABLE BOOKING (

BKGNO INT NOT NULL,

BKGDATE date NOT NULL,

BAL DECIMAL(8, 2),

TOTPRICE DECIMAL(8, 2),

PAIDAMT DECIMAL(8, 2) NOT NULL,

FPRICE DECIMAL(8, 2) NOT NULL,


DEPTTIME VARCHAR(30) NOT NULL,

ARRTIME VARCHAR(30) NOT NULL,

DEST CHAR(3) NOT NULL,

ORIG CHAR(3) NOT NULL,

BKGCITY INT NOT NULL,


PAIDBY INT NOT NULL,

CLASSID INT NOT NULL,

STATUSID INT NOT NULL,

PRIMARY KEY (BKGNO),

FOREIGN KEY (FNO, ORIG, DEST, DEPTTIME, ARRTIME) REFERENCES

FLIGHT_AVAILABILITY (FNO, ORIG, DEST, DEPTTIME, ARRTIME),

FOREIGN KEY (BKGCITY) REFERENCES CITY (CITYID),

FOREIGN KEY (CUSTID) REFERENCES CUSTOMER (CUSTID),

FOREIGN KEY (PAIDBY) REFERENCES CUSTOMER (CUSTID),

FOREIGN KEY (CLASSID) REFERENCES CLASS (CLASSID),

FOREIGN KEY (STATUSID) REFERENCES STATUS (STATUSID)

)

4.5 Airline Reservation Interactive Queries

Query 1: Give all the customers who lives in Canada and sort by customer_id.

SELECT CUSTID AS CUSTOMER_ID

FROM CUSTOMER

WHERE COUNTRY = 'Canada'

ORDER BY CUSTID

Query 2: List all different customers who made bookings.

SELECT DISTINCT CUSTID AS CUSTOMER_ID

FROM BOOKING

Query 3: Display all currency exchange rate is greater than 1. Please sort them by

from_currency and to_currency.

SELECT FCURR AS FROM_CURRENCY,TCURR AS TO_CURRENCY,EXCHRATE AS

EXCHANGE_RATE

FROM CURRENCY

WHERE EXCHRATE > 1


32

ORDER BY FCURR,TCURR

Query 4: List all the flight availabilities between Toronto (airport code is 'YYZ') and New

York (airport code is 'JFK'). Please display flight_no, origin, destinatin,

depature_time, and arrival_time. Please sort them by flight_no.

SELECT FNO AS FLIGHT_NO,ORIG AS ORIGIN,DEST AS DESTINATION,DEPTTIME

ASDEPATURE_TTIME,ARRTIME AS ARRIVAL_TIME

FROM FLIGHT_AVAILABILITY

WHERE ORIG = 'YYZ' AND DEST = 'JFK'

OR ORIG = 'JFK' AND DEST = 'YYZ'

ORDER BY FNO

Query 5: List all customers who did not place any booking. Please display customer_id

only, and sort records by customer_id.

SELECT CUSTID AS CUSTOMER_ID

FROM CUSTOMER

WHERE CUSTID NOT IN (SELECT DISTINCT CUSTID FROM BOOKING)

ORDER BY CUSTID

Query 6: Display all customer's first_name, last_name, phone_no (format like 416-111-

2222) and email. Please sort them by customer_id.

SELECT C.CUSTID AS CUSTOMER_ID,

C.FNAME AS FIRST_NAME,

C.LNAME AS LAST_NAME,

P.PCRTYCODE||'-'||P.PAREACODE||'-'||P.PNUMBER AS PHONE_NO,

E.EMAIL AS EMAIL

FROM CUSTOMER C

RIGHT JOIN PHONE P ON C.CUSTID = P.CUSTID

RIGHT JOIN EMAIL E ON C.CUSTID = E.CUSTID

ORDER BY C.CUSTID

Query 7: List all canceled bookngs. please display booking_no, customer_id, flight_no,

origin, destination, class, status, and booking_city. Please also sort by

booking_no, customer_id and flight_no.

SELECT BKGNO AS BOOKING_NO,CUSTID AS CUSTOMER_ID,FNO AS FLIGHT_NO,

ORIG AS ORIGIN,DEST AS DESTINATION,C.CLASS AS CLASS,

S.STATUS AS STATUS,CITY.CITYNM AS BOOKING_CITY

FROM BOOKING B

INNER JOIN STATUS S ON B.STATUSID = S.STATUSID

INNER JOIN CLASS C ON B.CLASSID = C.CLASSID

INNER JOIN CITY ON B.BKGCITY = CITY.CITYID

WHERE S.STATUS = 'Canceled'

ORDER BY BKGNO,CUSTID,FNO

Query 8: List total_price, total_payment and total_balance for each city. Please exclude

canceled bookings and sort records by city_name.


33

SELECT C.CITYNM AS CITY,SUM(TOTPRICE)AS TOTAL_PRICE,

SUM(PAIDAMT) AS TOTAL_PAYMENT,SUM(BAL) AS TOTAL_BALANCE

FROM BOOKING B

INNER JOIN CITY C ON B.BKGCITY = C.CITYID

WHERE STATUSID <> 2

GROUP BY C.CITYNM

ORDER BY C.CITYNM

Query 9: Calculate new total_price for each booking if origin airport tax increase by 0.01

and destination airport tax decrease by 0.005. Please display booking_no,

origin, destination, flight_price, previous_total_price and new_total_price.

SELECT BKGNO AS BOOKING_NO,ORIG AS ORIGIN,

DEST AS DESTINATION,FPRICE AS FLIGHT_PRICE,

TOTPRICE AS PREVIOUS_TOTAL_PRICE,

FPRICE*(1+(O.AIRPORTTAX+0.01)+(D.AIRPORTTAX-0.005)) AS

NEW_TOTAL_PRICE

FROM BOOKING B

INNER JOIN AIRPORT O ON B.ORIG = O.AIRPORTCD

INNER JOIN AIRPORT D ON B.DEST = D.AIRPORTCD

Query 10: List number_of_bookings, number_of_emails, number_of_phones and

mumber_of_faxs for each customer.

SELECT C.CUSTID AS CUSTOMER_ID

,(SELECT COUNT(CUSTID) FROM BOOKING WHERE CUSTID=C.CUSTID) AS

NUMBER_OF_BOOKINGS

,(SELECT COUNT(CUSTID) FROM EMAIL WHERE CUSTID=C.CUSTID) AS

NUMBER_OF_EMAILS

,(SELECT COUNT(CUSTID) FROM PHONE WHERE CUSTID=C.CUSTID) AS

NUMBER_OF_PHONES

,(SELECT COUNT(CUSTID) FROM FAX WHERE CUSTID=C.CUSTID) AS

NUMBER_OF_FAXS

FROM CUSTOMER C


34

Chapter 5: Movie Rental

5.1 Movie Rental Informal Description

The movies are rented out in stores and there are several stores. Each store has a unique

distributor that supplies the store with tapes. A distributor may supply more than one

store. Each distributor has a name, an address, and a phone number. Each store has a

name, an address, and a phone number. For each employee we must keep the following

information: working store, a name, a supervisor, an address , a phone number, SIN

(social insurance number) and the date when the employee was hired. For each customer

we have to keep the following information: a name, an address, and a phone number (if

any).

For each rental, we must keep track of which employee served the customer,

which movie and which copy (i.e. type) the customer rented, information about payment,

the date and the time of the rental, the status (rented, returned_in_time, returned_late), the

rate (i.e. the price), and if applicable, due date and overdue charges. About the payment

we have to keep which of the employees accepted the payment (does not have to be the

same employee who rented the tape), the type of payment (i.e. cash, check, credit card,

direct debit – for each type you must provide for relevant information to be kept, e.g.

credit card number if credit card is used), the amount of the payment, date + time of the

payment, payment status (completed if cash or the money have been received, approved if

debit or credit card go through, pending if the check has not cleared yet). About each tape

we have to keep information in what condition the tape is and what movie is on the tape.

About each movie we have to keep its title, director’s name, simple description, the name

of a (single) major star, the movie’s rating (use numbers 1-5).


35

5.2 Movie Rental Logical Model

Figure 6: The Movie Rental Logical Model


36

5.3 Movie Rental Physical DB2 Model

Figure 7: The Movie Rental Physical Model


37

5.4 Movie Rental DB2 Schema

CREATE TABLE DISTRIBUTOR(

DISTID INTEGER NOT NULL,

DISTNM VARCHAR(20),

DISTSTR VARCHAR(30),

DISTCITY VARCHAR(20),

DISTPRVN CHAR(2),

DISTPC CHAR(6),

DISTPNO CHAR(12),

PRIMARY KEY (DISTID)

)

CREATE TABLE STORE(

STORENM VARCHAR(20),

STORESTR VARCHAR(30),

STORECITY VARCHAR(20),

STOREPVN CHAR(2),

STOREPC CHAR(6),

STOREPNO CHAR(12),

STOREID INTEGER NOT NULL,

DISTID INTEGER NOT NULL,

PRIMARY KEY (STOREID),

FOREIGN KEY (DISTID) REFERENCES DISTRIBUTOR (DISTID)

)

CREATE TABLE EMPLOYEE(

EMPSIN CHAR(9) NOT NULL,

MGRSIN CHAR(9),

EMPNM VARCHAR(20),

EMPSTR VARCHAR(30),

EMPCITY VARCHAR(20),

EMPPVN CHAR(2),

EMPPC CHAR(6),

EMPPNO CHAR(12),

EMPHIREDT DATE,


PRIMARY KEY (EMPSIN),

FOREIGN KEY (MGRSIN) REFERENCES EMPLOYEE (EMPSIN),

FOREIGN KEY (STOREID) REFERENCES STORE (STOREID)

)

CREATE TABLE CUSTOMER(

CUSTID INTEGER NOT NULL,

CUSTNM VARCHAR(20),

CUSTSTR VARCHAR(30),

CUSTCITY VARCHAR(20),

CUSTPVN CHAR(2),

CUSTPC CHAR(6),

CUSTPNO CHAR(12),

PRIMARY KEY (CUSTID)

)

CREATE TABLE MOVIE(

MOVIETITL VARCHAR(20),

MOVIEDESC VARCHAR(50),


38

DIRECTOR VARCHAR(20),

RATING INTEGER,

STARNM VARCHAR(20),

MOVIEID INTEGER NOT NULL,

PRIMARY KEY (MOVIEID)

)

CREATE TABLE STORE_MOVIE(



PRIMARY KEY (MOVIEID, STOREID),

FOREIGN KEY (MOVIEID) REFERENCES MOVIE (MOVIEID),

FOREIGN KEY (STOREID) REFERENCES STORE (STOREID)

)

CREATE TABLE TAPE(

TAPEID INTEGER NOT NULL,

CONDITION VARCHAR(20),



PRIMARY KEY (TAPEID, MOVIEID, STOREID),

FOREIGN KEY (MOVIEID, STOREID) REFERENCES STORE_MOVIE (MOVIEID,

STOREID)

)

CREATE TABLE PAYMENT_STATUS(

PSTATUSID INTEGER NOT NULL,

PDESC VARCHAR(20),

PRIMARY KEY (PSTATUSID)

)

CREATE TABLE PAYMENT_TYPE(

PTID INTEGER NOT NULL,

PTDESC VARCHAR(10),

PRIMARY KEY (PTID)

)

CREATE TABLE PAYMENT(

PAYID INTEGER NOT NULL,

AMT DECIMAL(8, 2),

PAYDTTM TIMESTAMP,



PSTATUSID INTEGER NOT NULL,

PTID INTEGER NOT NULL,

PRIMARY KEY (PAYID),

FOREIGN KEY (EMPSIN) REFERENCES EMPLOYEE (EMPSIN),


FOREIGN KEY (PSTATUSID) REFERENCES PAYMENT_STATUS (PSTATUSID),

FOREIGN KEY (PTID) REFERENCES PAYMENT_TYPE (PTID)

)

CREATE TABLE RENTAL_STATUS(

RSTATUSID INTEGER NOT NULL,

RDESC VARCHAR(20),

PRIMARY KEY (RSTATUSID)

)

CREATE TABLE MOVIE_RENTAL(

DUEDT DATE,

OVERDUECRG DECIMAL(8, 2),


39

RDTTM TIMESTAMP NOT NULL,



TAPEID INTEGER NOT NULL,

RRATE DECIMAL(8, 2),

PAYID INTEGER,



RSTATUSID INTEGER NOT NULL,

PRIMARY KEY (TAPEID, MOVIEID, STOREID, RDTTM),

FOREIGN KEY (EMPSIN) REFERENCES EMPLOYEE (EMPSIN),


FOREIGN KEY (TAPEID, MOVIEID, STOREID) REFERENCES TAPE (TAPEID,

MOVIEID, STOREID),

FOREIGN KEY (PAYID) REFERENCES PAYMENT (PAYID),

FOREIGN KEY (RSTATUSID) REFERENCES RENTAL_STATUS (RSTATUSID)

)

CREATE TABLE DEBIT_CARD(

DNO INTEGER,

DTYPE VARCHAR(20),

DEXPR CHAR(5),



FOREIGN KEY (PAYID) REFERENCES PAYMENT (PAYID)

)

CREATE TABLE CREDIT_CARD(

CNO INTEGER,

CTYPE VARCHAR(20),

CEXPR CHAR(5),




)

CREATE TABLE CHECK(

CHECKNO VARCHAR(30),

BANKNO VARCHAR(20),

BANKNM VARCHAR(30),




)

5.5 Movie Rental Interactive Queries

Query 1: Give all the customers who lives in Hamilton. Display customer_id and

customer_name.

SELECT CUSTID AS CUSTOMER_ID,CUSTNM AS CUSTOMER_NAME

FROM CUSTOMER

WHERE CUSTCITY='Hamilton'


40

Query 2: Display the total payment are received by each employee, and sort by empsin.

SELECT EMPSIN AS EMPLOYEE_SIN,SUM(AMT) AS TOTAL_AMT

FROM PAYMENT

GROUP BY EMPSIN

ORDER BY EMPSIN

Query 3: Display the total movies are rented out by each store, and sort by storeid.

SELECT STOREID,COUNT(TAPEID) AS MOVIE_RENTED

FROM MOVIE_RENTAL

GROUP BY STOREID

ORDER BY STOREID

Query 4: Display all the tapes are never rented out in every store, and sort by movieid

& tapeid.

SELECT DISTINCT T.MOVIEID,T.TAPEID

FROM TAPE T

LEFT JOIN MOVIE_RENTAL R ON T.MOVIEID=R.MOVIEID AND T.TAPEID=R.TAPEID

WHERE R.MOVIEID IS NULL

ORDER BY T.MOVIEID,T.TAPEID

Query 5: Display all customers who did not rent any movie so far and sort by custid.

SELECT CUSTID,CUSTNM

FROM CUSTOMER

WHERE CUSTID NOT IN (SELECT DISTINCT CUSTID FROM MOVIE_RENTAL)

ORDER BY CUSTID

Query 6: Display the total amount received by different payment type, and sort by

ptdesc.

SELECT TP.PTDESC AS PAYMENT_TYPE,SUM(AMT) AS TOTAL_AMT

FROM PAYMENT P

INNER JOIN PAYMENT_TYPE TP ON P.PTID = TP.PTID

GROUP BY TP.PTDESC

ORDER BY TP.PTDESC

Query 7. Display the number of movies rented out based on the movie rating, and sort by

rating.

SELECT M.RATING,COUNT(MR.MOVIEID) AS NO_OF_MOVIES

FROM MOVIE_RENTAL MR

INNER JOIN MOVIE M ON MR.MOVIEID = M.MOVIEID

GROUP BY M.RATING

Query 8: Display top 5 customers based on their total payment, and sort their total

payment decreased.

SELECT CUSTID,SUM(AMT) AS TOTAL_AMT


41

FROM PAYMENT

GROUP BY CUSTID

ORDER BY TOTAL_AMT DESC

FETCH FIRST 5 ROWS ONLY

Query 9: List all the movies customer rented. Please display the columns: movie_title,

rental_status, rental_rate, rental_employee, the employ accept the

payment, payment_type and payment_status.

SELECT M.MOVIETITL AS MOVIE_TITLE,

RS.RDESC AS RENTAL_STATUS,

MR.RRATE AS RENTAL_RATE,

E1.EMPNM AS RENTAL_EMPLOYEE,

E2.EMPNM AS CASHIER_EMPLOYEE,

PT.PTDESC AS PAYMENT_TYPE,

PS.PDESC AS PAYMENT_STATUS

FROM MOVIE_RENTAL MR

INNER JOIN CUSTOMER C ON MR.CUSTID = C.CUSTID

INNER JOIN MOVIE M ON MR.MOVIEID = M.MOVIEID

INNER JOIN RENTAL_STATUS RS ON MR.RSTATUSID = RS.RSTATUSID

INNER JOIN EMPLOYEE E1 ON MR.EMPSIN = E1.EMPSIN

LEFT JOIN PAYMENT P ON MR.PAYID = P.PAYID

LEFT JOIN EMPLOYEE E2 ON P.EMPSIN = E2.EMPSIN

LEFT JOIN PAYMENT_STATUS PS ON P.PSTATUSID = PS.PSTATUSID

LEFT JOIN PAYMENT_TYPE PT ON P.PTID = PT.PTID

WHERE C.CUSTNM = 'customer1'

Query 10: List all the manager's name and the name of employee they manage.

Please sort by manager sin & employee sin.

SELECT MGR.EMPNM AS MANAGER_NAME,E.EMPNM AS EMPLOYEE_NAME

FROM EMPLOYEE E

INNER JOIN EMPLOYEE MGR ON E.MGRSIN = MGR.EMPSIN

ORDER BY MGR.EMPSIN,E.EMPSIN


42

Chapter 6: Car Rental

6.1 Car Rental Informal Description

Our company does car rental business and has several locations with different address

(address consist of street or rural route with the number, city, province and postal code).

The cars are classified as subcompacts, compacts, sedans, or luxury. Each car has a

particular make, model, year made, and color. Each car has a unique identification

number and a unique license plate.

The cars rented in a particular location may be returned to a different location (so-

called drop off). For every car we keep the odometer reading before it is rented and after

it is returned. Since we trust our customers, we do not record the defect when the car is

rent out and returned back. However, we rent the car with full tank and record the volume

of gas in the tank when the car is returned, but we only indicate if the tank is empty,

quarter full, half full, three quarters full, or full.

We keep track of which day a car was rented, but not of the time, similarly for car

returning. If a customer requests a specific class (say sedan), we may rent the customer a

higher-class car if we do not have the requested class in the stock, but we will price it at

the level the customer requested (so-called upgrade). Each car class has its own pricing,

but all cars in the same class are priced the same. We have rental policies for 1 day, 1

week, 2 weeks, and 1 month. Thus, if a customer rents a car for 8 days, it will be priced as

1 week + 1 day. The drop-off charge only depends on the class of the rented car, the

location it was rented from and the location it is returned to.

About our customers, we keep their names, addresses, possibly all phone

numbers, and the number of the driver’s license (we assume a unique license per person).

About our employees we keep the same information (we require that all our employees

have a driver’s license). We have several categories of workers, drivers, cleaners, clerks,

and managers. Any of our employees can rent a car from our company for a 50%

discount, if the rental is less than 2 weeks. However, for any longer rental they must pay

90% of the regular price. Every employee works in one location only. We have

headquarters in Hamilton. The people who work there are all classified as managers, one

of them is the president, two of them are the vice-presidents, one for operation, the other

for marketing).


43

For certain weeks we have promotional rentals that are usually 60% of the regular

price, but may be also of different percentage. They always affect only a single class of

cars – i.e. we may have a promotion for subcompacts, but during that week we do not

have any promotions for compacts, sedans or luxury cars. During some years we can have

many promotions, in some we have none. The promotions cannot be applied to the

employees.


44

6.2 Car Rental Logical Model

Figure 8: The Car Rental Logic Model


45

6.3 Car Rental Physical DB2 Model

Figure 9: The Car Rental Physical Model


46

6.4 Car Rental DB2 Schema

CREATE TABLE ADDRESS (

ADDRESSID INTEGER NOT NULL,

SNUMBER CHAR(10) NOT NULL,

STREET CHAR(100) NOT NULL,

CITY CHAR(20) NOT NULL,

PROVINCE CHAR(2) NOT NULL,

PCODE CHAR(6) NOT NULL,

IS_HQUARTERS CHAR(1) NOT NULL,

PRIMARY KEY (ADDRESSID),

UNIQUE (SNUMBER,STREET,CITY,PROVINCE),

CHECK(ADDRESSID > 0),

CHECK(PROVINCE IN ('AL', 'BC','MA','NB','NL','NT','NS','NU',

'ON','PE','QU','SA','YT')),

CHECK (IS_HQUARTERS IN ('H','0'))

)

CREATE TABLE CLASSTABLE (

CLASS CHAR(1) NOT NULL,

CDESC VARCHAR(20) NOT NULL,

PRIMARY KEY (CLASS),

CHECK (CLASS IN ('B','C','S','L'))

)

CREATE TABLE CAR (

CARID INTEGER NOT NULL,

MAKE CHAR(10) NOT NULL,

MODEL CHAR(20) NOT NULL,

YEAR CHAR(4) NOT NULL,

COLOUR CHAR(10) NOT NULL,

LPLATE CHAR(8) NOT NULL,

CLOCATION INTEGER,

CLASS CHAR(1),

PRIMARY KEY (carid),

FOREIGN KEY (CLOCATION) REFERENCES ADDRESS (ADDRESSID),

FOREIGN KEY (CLASS) REFERENCES CLASSTABLE,

CHECK (CARID > 0)

)

CREATE TABLE PERSON (

DLICENSE CHAR(20) NOT NULL,

FNAME CHAR(50) NOT NULL,

LNAME CHAR(50) NOT NULL,

ADDRESSID INTEGER NOT NULL,

PRIMARY KEY (DLICENSE),

FOREIGN KEY (ADDRESSID) REFERENCES ADDRESS

)

CREATE TABLE PHONE (


PHONE CHAR(20) NOT NULL,

PRIMARY KEY (DLICENSE,PHONE),

FOREIGN KEY (DLICENSE) REFERENCES PERSON

)

CREATE TABLE ETYPETABLE (

ETYPE CHAR(1) NOT NULL,


47

PRIMARY KEY (ETYPE),

CHECK (ETYPE IN ('D', 'C', 'K', 'M'))

)

CREATE TABLE EMPLOYEE (


LOCATION INTEGER NOT NULL,

ETYPE CHAR(1) NOT NULL,

IS_PRES CHAR(1) NOT NULL,

IS_VIP CHAR(1) NOT NULL,


FOREIGN KEY (DLICENSE) REFERENCES PERSON,

FOREIGN KEY (LOCATION) REFERENCES ADDRESS (ADDRESSID),

FOREIGN KEY (ETYPE) REFERENCES ETYPETABLE,

CHECK (IS_PRES IN ('P','0')),

CHECK (IS_VIP IN ('M','P','0')),

CHECK ((IS_PRES='P' AND ETYPE='M') OR

(IS_VIP='M' AND ETYPE='M') OR

(IS_VIP='P' AND ETYPE='M') OR

(IS_VIP='0' AND IS_PRES='0')),

CHECK ((IS_PRES='P' AND IS_VIP='0') OR

(IS_PRES='0' AND IS_VIP='M') OR

(IS_PRES='0' AND IS_VIP='P') OR

(IS_PRES='0' AND IS_VIP='0'))

)

CREATE TABLE CUSTOMER (



FOREIGN KEY (DLICENSE) REFERENCES PERSON

)

CREATE TABLE PRATE (

DURATION CHAR(1) NOT NULL,


AMOUNT DECIMAL(8,2) NOT NULL,

PRIMARY KEY (DURATION,CLASS),


CHECK (DURATION IN ('D', 'W', 'T', 'M'))

)

CREATE TABLE PROMO (



FROM_DATE DATE NOT NULL,

PERCENTAGE DECIMAL(4,2) NOT NULL,

PRIMARY KEY (DURATION,CLASS),

FOREIGN KEY (DURATION,CLASS) REFERENCES PRATE

)

CREATE TABLE RENTAL (

RENTALID INTEGER NOT NULL,


CARID INTEGER NOT NULL,

FROM_LOCATION INTEGER NOT NULL,

DROPOFF_LOCATION INTEGER NOT NULL,

FROM_DATE DATE NOT NULL,

TO_DATE DATE,

TANK CHAR(1) NOT NULL,


48

INIT_ODO INTEGER NOT NULL,

RETURN_ODO INTEGER,

PRIMARY KEY (RENTALID),

UNIQUE (CARID,FROM_DATE),

FOREIGN KEY (CARID) REFERENCES CAR,

FOREIGN KEY (DLICENSE) REFERENCES CUSTOMER,

FOREIGN KEY (FROM_LOCATION) REFERENCES ADDRESS (ADDRESSID),

FOREIGN KEY (DROPOFF_LOCATION) REFERENCES ADDRESS (ADDRESSID),

CHECK (RENTALID > 0),

CHECK (TANK IN ('F','T','H','Q','E')),

CHECK ((FROM_DATE < TO_DATE) OR (TO_DATE IS NULL)),

CHECK ((INIT_ODO < RETURN_ODO) OR (RETURN_ODO IS NULL))

)

CREATE TABLE RENTALRATE (

RENTALID INTEGER NOT NULL,



COUNT INTEGER NOT NULL,

PRIMARY KEY (RENTALID,DURATION,CLASS),

FOREIGN KEY (RENTALID) REFERENCES RENTAL,

FOREIGN KEY (DURATION,CLASS) REFERENCES PRATE,

CHECK (COUNT > 0)

)

CREATE TABLE DROPOFFCHARGE (


FROM_LOCATION INTEGER NOT NULL,

TO_LOCATION INTEGER NOT NULL,

CHARGE DECIMAL(4,2) NOT NULL,

PRIMARY KEY (CLASS,FROM_LOCATION,TO_LOCATION),


FOREIGN KEY (FROM_LOCATION) REFERENCES ADDRESS (ADDRESSID),

FOREIGN KEY (TO_LOCATION) REFERENCES ADDRESS (ADDRESSID)

)

6.5 Car Rental Interactive Queries

Query 1: Give last name of all customers who are now renting a car from our company.

SELECT LNAME

FROM CUSTOMER, PERSON, RENTAL

WHERE CUSTOMER.DLICENSE=RENTAL.DLICENSE AND CUSTOMER.DLICENSE =

PERSON.DLICENSE AND TO_DATE IS NULL

Query 2: Give make and color of all cars currently rented out.

SELECT MAKE,COLOUR

FROM CAR,RENTAL

WHERE CAR.CARID=RENTAL.CARID AND TO_DATE IS NULL

Query 3: For each completed rental, give the rental price and rental_id.


49

SELECT SUM(CHARGE), RENTALID

FROM (SELECT RENTALID,AMOUNT*COUNT

FROM RENTALRATE,PRATE

WHERE RENTALRATE.DURATION=PRATE.DURATION

AND RENTALRATE.CLASS=PRATE.CLASS) AS T(RENTALID,CHARGE)

GROUP BY RENTALID

Query 4: List last name of all managers.

SELECT LNAME

FROM EMPLOYEE, PERSON

WHERE ETYPE='M' AND EMPLOYEE.DLICENSE=PERSON.DLICENSE

Query 5: List last and first names of all customers.

SELECT LNAME, FNAME

FROM CUSTOMER, PERSON

WHERE CUSTOMER.DLICENSE=PERSON.DLICENSE

Query 6: Give a query that answers the question "Is any of our employee also our

customer"?

SELECT *

FROM EMPLOYEE, CUSTOMER

WHERE EMPLOYEE.DLICENSE=CUSTOMER.DLICENSE

Query 7: Does our president work in the headquarters?

SELECT *

FROM EMPLOYEE, ADDRESS

WHERE IS_PRES='P' AND LOCATION=ADDRESSID AND ADDRESS.IS_HQUARTERS='H'

Query 8: Find rental_id of all shortest (completed) rentals.

SELECT RENTALID

FROM RENTAL

WHERE TO_DATE IS NOT NULL AND (DAYS(TO_DATE)-DAYS(FROM_DATE)) IN

(SELECT MIN(DAYS(TO_DATE) - DAYS(FROM_DATE))

FROM RENTAL

WHERE TO_DATE IS NOT NULL)

Query 9: Find the value of the cheapest (completed) rental. We will utilize query 3 as the

inner query.

SELECT MIN(CHARGE)

FROM (SELECT SUM(CHARGE), RENTALID

FROM (SELECT RENTALID,AMOUNT*COUNT

FROM RENTALRATE,PRATE

WHERE RENTALRATE.DURATION=PRATE.DURATION

AND RENTALRATE.CLASS=PRATE.CLASS) AS T(RENTALID,CHARGE)


50

GROUP BY RENTALID) AS T1(CHARGE,RENTALID)

Query 10: Give makes of the cars that have never been rented.

SELECT DISTINCT MAKE FROM CAR

EXCEPT

SELECT DISTINCT MAKE FROM CAR, RENTAL WHERE CAR.CARID=RENTAL.CARID


51

Chapter 7: Course Registration

7.1 Course Registration Informal Description

In our college, students need to register for courses before new semester start. All of

programs in our college take four years to finish.

Each course has a unique designation, title, description, year (in which year of

study the course is to be taken, for instance 2nd

year course), and classroom. A course can

have no or many tutorial sections, and no or many lab sections. Each course is taught by

exactly one instructor. Each instructor has a unique id, name, departmental affiliation,

office room, phone extension, and a unique email address. Each student has a unique id,

name, and the year of his/her study. A student cannot be an instructor. A course can have

zero or many tutorial sections unique to the course (i.e. tutorial sections are not shared by

different courses). Each tutorial section has exactly one TA assigned. A TA can tutor

more than one tutorial section for the same course, and any number of tutorials for

different courses. A TA cannot be an instructor, however a student can work as a TA (in

that case his/her student id is used as TA id). A course can have zero or many lab sections

unique to the course (i.e. lab sections are not shared by different courses). Each lab

section has exactly one LA assigned. An LA can oversee more than one lab section for

the same course, and any number of labs for different courses. An LA cannot be an

instructor, however a student can work as a LA (in which case his/her student id is used

as the LA id). In fact, a student can work as a TA and an LA simultaneously. Thus, a TA

may or may not be a student, an LA may or may not be a student. A person can work as

both, a TA and a LA. TA has the same attributes as instructor, the same goes for LA.

Each course has zero to many courses designated as its prerequisites and zero to

many courses designated as its anti-requisites. Prerequisite courses are of the same or

lower year, anti-requisite courses are of the same year. In the system we keep information

of what courses a student has taken and what courses the student is registering. All

courses are either Pass or Fail. A student can register a course only if he/she has passed

all the prerequisites and has not passed any or is not registered in any of the anti-

requisites. A student can only register a course of the appropriate year, i.e. a student in

year X of study can only register and take course of year X.


52

7.2 Course Registration Logical Model

Figure 10:The Course Registration Logical Model


53

7.3 Course Registration Physical DB2 Model

Figure 11: The Course Registration Physical Model


54

7.4 Course Registration DB2 Schema

CREATE TABLE COURSE (

DESIG CHAR(5) NOT NULL,

TITLE CHAR(30) NOT NULL,

DESCR CHAR(150) NOT NULL,

CLASSROOM CHAR(5) NOT NULL,

PRIMARY KEY(DESIG)

)

CREATE TABLE Y1COURSE (


PRIMARY KEY(DESIG)

)



PRIMARY KEY(DESIG)

)



PRIMARY KEY(DESIG)

)



PRIMARY KEY(DESIG)

)

CREATE TABLE PREREQ11 (

DESIG1 CHAR(5) NOT NULL,


PRIMARY KEY(DESIG1,DESIG2)

)




PRIMARY KEY(DESIG1,DESIG2),

FOREIGN KEY (DESIG1) REFERENCES Y1COURSE (DESIG),


CONSTRAINT PREREQ12_C3 CHECK (DESIG1<>DESIG2)

)








)








55


)








)








)








)








)








)








)

CREATE TABLE ANTIREQ1 (





56



CONSTRAINT ANTIREQ1_C3 CHECK (DESIG1<>DESIG2)

)








)








)








)

CREATE TABLE PERSON (

ID CHAR(7) NOT NULL,

NAME CHAR(20) NOT NULL,

PRIMARY KEY(ID)

)

CREATE TABLE STUDENT (


YEAR INTEGER NOT NULL,

PRIMARY KEY(ID),

FOREIGN KEY (ID) REFERENCES PERSON (ID)

)

CREATE TABLE INSTRUCTOR (


DEPT CHAR(4) NOT NULL,

ROOM CHAR(5) NOT NULL,

EXTENSION CHAR(5) NOT NULL,

EMAIL CHAR(20) NOT NULL,

PRIMARY KEY(ID),

FOREIGN KEY (ID) REFERENCES PERSON (ID),

CONSTRAINT INSTRUCTOR_C2 UNIQUE (EMAIL)

)

CREATE TABLE STAFF (


DEPT CHAR(4) NOT NULL,

ROOM CHAR(5) NOT NULL,

EXTENSION CHAR(5) NOT NULL,


57

EMAIL CHAR(20) NOT NULL,

PRIMARY KEY(ID),


CONSTRAINT STAFF_C2 UNIQUE (EMAIL)

)

CREATE TABLE LAB (


SECTION INTEGER NOT NULL,

LABROOM CHAR(5) NOT NULL,

PRIMARY KEY(DESIG,SECTION),

FOREIGN KEY (DESIG) REFERENCES COURSE (DESIG)

)

CREATE TABLE TUTORIAL (



CLASSROOM CHAR(5) NOT NULL,


FOREIGN KEY (DESIG) REFERENCES COURSE (DESIG)

)

CREATE TABLE TA (


PRIMARY KEY(ID),

FOREIGN KEY (ID) REFERENCES STAFF (ID)

)

CREATE TABLE LA (


PRIMARY KEY(ID),

FOREIGN KEY (ID) REFERENCES STAFF (ID)

)

CREATE TABLE HASTA (





FOREIGN KEY (DESIG,SECTION) REFERENCES TUTORIAL (DESIG,SECTION),

FOREIGN KEY (ID) REFERENCES TA (ID)

)

CREATE TABLE HASLA (





FOREIGN KEY (DESIG,SECTION) REFERENCES LAB (DESIG,SECTION),

FOREIGN KEY (ID) REFERENCES LA (ID)

)

CREATE TABLE HASI (



PRIMARY KEY(DESIG),

FOREIGN KEY (DESIG) REFERENCES COURSE (DESIG),

FOREIGN KEY (ID) REFERENCES INSTRUCTOR (ID)

)

CREATE TABLE L1 (



58


STATUS CHAR(1) NOT NULL CHECK (STATUS IN ('P','F','R')),

PRIMARY KEY(ID,DESIG),

FOREIGN KEY (ID) REFERENCES STUDENT (ID),

FOREIGN KEY (DESIG) REFERENCES Y1COURSE (DESIG)

)

CREATE TABLE L2 (







)

CREATE TABLE L3 (







)

CREATE TABLE L4 (







)

7.5 Course Registration Interactive Queries

Query 1: List all triples (student name, course designation, status) of courses taken or

registered by students with id '0000041' and '0000042'. status is the status of

each course (i.e. 'P' for passed, 'F' for failed, 'R' for registered).

SELECT NAME,DESIG,STATUS

FROM STUDENT, PERSON,L1

WHERE STUDENT.ID=L1.ID

AND (STUDENT.ID='0000041' OR STUDENT.ID='0000042')

AND STUDENT.ID=PERSON.ID

UNION

SELECT NAME,DESIG,STATUS

FROM STUDENT, PERSON,L2

WHERE STUDENT.ID=L2.ID

AND (STUDENT.ID='0000041' OR STUDENT.ID='0000042')

AND STUDENT.ID=PERSON.ID

Query 2: List names of all students in year 1.

SELECT PERSON.NAME


59

FROM PERSON, STUDENT

WHERE PERSON.ID=STUDENT.ID AND STUDENT.YEAR=1

Query 3: List names of all instructors teaching year 1 courses.

SELECT PERSON.NAME

FROM PERSON, INSTRUCTOR, HASI, Y1COURSE

WHERE PERSON.ID=INSTRUCTOR.ID

AND HASI.ID=INSTRUCTOR.ID AND HASI.DESIG=Y1COURSE.DESIG

Query 4: List designation of all courses that have tutorials. No designation can repeat.

SELECT DISTINCT DESIG FROM TUTORIAL

Query 5: List designation of all courses that have labs with more than 1 section. No

designation can repeat.

SELECT DISTINCT DESIG FROM LAB WHERE SECTION=2

Query 6: List names of instructors that teach at least one course with multiple sections

labs. No name can repeat.

SELECT DISTINCT NAME

FROM PERSON,INSTRUCTOR,HASI

WHERE (INSTRUCTOR.ID=PERSON.ID) AND (HASI.ID=PERSON.ID)

AND HASI.DESIG IN (SELECT COURSE.DESIG FROM COURSE,LAB

WHERE COURSE.DESIG=LAB.DESIG AND SECTION=2)

Query 7: List names of instructors that teach only courses with single-sections labs or no

labs. No name can repeat.

SELECT DISTINCT NAME

FROM PERSON, INSTRUCTOR

WHERE (PERSON.ID=INSTRUCTOR.ID)

AND NAME NOT IN (SELECT DISTINCT NAME FROM PERSON,INSTRUCTOR,HASI

WHERE (INSTRUCTOR.ID=PERSON.ID) AND (HASI.ID=PERSON.ID)

AND HASI.DESIG IN (SELECT COURSE.DESIG FROM COURSE,LAB

WHERE COURSE.DESIG=LAB.DESIG

AND SECTION=2))


60

Chapter 8: Emergency Room

8.1 Emergency Room Informal Description

In our Emergency Room (ER), we have three distinct types of workers: receptionists,

nurses, and doctors. Any of the workers can in fact be a patient. Each person in the

proposed system, be it a patient or a worker has a last, a first, possibly a middle name, and

one or more addresses. An address consists of a country, province, city, street and street

number. Each person can have none or more email addresses, none or more telephone

numbers.

The workers work in ER in shifts. A shift consists of start and end time. The shifts

do not overlap, but they are consecutive, i.e. there is a shift on at any given time and day.

We are assuming that the model we are creating (and eventually the database we will

design) covers some extended period of time. Each worker will thus be assigned to many

shifts in that period. Exactly two receptionists are assigned to each shift, a group of two or

more nurses is assigned to each shift, a group of two or more doctors is assigned to each

shift, one of the doctors assigned to a shift is the shift’s triage doctor.

When a patient comes to ER, it happens during a particular shift. The patient is

admitted by a particular receptionist, is seen by the triage doctor of the shift. The patient

may be send home, prescribed some medication by the triage doctor and send home, or is

staying in ER – in which case the patient is assigned a bed and case doctors (one of the

doctors on each shift best qualified for the particular problem of the patient). Each bed is

supervised by a single nurse during a shift, but a nurse may supervise many beds, or none

at all. The case doctor(s) may prescribe a medication that is administered to the patient by

a single nurse in each shift for the duration of the patient taking the medicine. Each

medication has a name, and for each patient there may be a different dosage and different

number of times a day to take it.


61

8.2 Emergency Room Logical Model

Figure 12: The Emergency Room Logical Model


62

8.3 Emergency Room Physical DB2 Model

Figure 13: The Emergency Room Physical Model


63

8.4 Emergency Room DB2 Schema

CREATE TABLE PERSON(

ID INTEGER NOT NULL,

LASTNAME CHAR(10) NOT NULL,

FIRSTNAME CHAR(10) NOT NULL,

MIDDLENAME CHAR(10),

PRIMARY KEY (ID)

)

CREATE TABLE PATIENT(

PID INTEGER NOT NULL,

PRIMARY KEY (PID),

FOREIGN KEY (PID) REFERENCES PERSON (ID)

)

CREATE TABLE WORKER(

WID INTEGER NOT NULL,

PRIMARY KEY (WID),

FOREIGN KEY (WID) REFERENCES PERSON (ID)

)

CREATE TABLE RECEPTIONIST(

RID INTEGER NOT NULL,

PRIMARY KEY (RID),

FOREIGN KEY (RID) REFERENCES WORKER (WID)

)

CREATE TABLE NURSE(

NID INTEGER NOT NULL,

PRIMARY KEY (NID),

FOREIGN KEY (NID) REFERENCES WORKER (WID)

)

CREATE TABLE DOCTOR(

DID INTEGER NOT NULL,

PRIMARY KEY (DID),

FOREIGN KEY (DID) REFERENCES WORKER (WID)

)

CREATE TABLE EMAIL(

EADDRESS CHAR(20) NOT NULL,

PRIMARY KEY (EADDRESS)

)

CREATE TABLE PHONENO(

AREACODE CHAR(3) NOT NULL,

NUMBER CHAR(7) NOT NULL,

PRIMARY KEY (AREACODE,NUMBER)

)

CREATE TABLE ADDRESS(




STREETNO CHAR(6) NOT NULL,

PRIMARY KEY (PROVINCE,CITY,STREET,STREETNO)

)

CREATE TABLE MEDICATION(


PRIMARY KEY (NAME)


64

)

CREATE TABLE BED(


PRIMARY KEY (NUMBER)

)

CREATE TABLE SHIFT(

SHIFTID INTEGER NOT NULL,

FROM TIMESTAMP NOT NULL,

TO TIMESTAMP NOT NULL,

PRIMARY KEY (SHIFTID),

UNIQUE (FROM,TO),

CHECK (FROM < TO)

)

CREATE TABLE HASE(


EADDRESS CHAR(20) NOT NULL,

PRIMARY KEY (ID,EADDRESS),


FOREIGN KEY (EADDRESS) REFERENCES EMAIL (EADDRESS)

)

CREATE TABLE HASP(


AREACODE CHAR(3) NOT NULL,


PRIMARY KEY (ID,AREACODE,NUMBER),


FOREIGN KEY (AREACODE,NUMBER) REFERENCES PHONENO(AREACODE,NUMBER)

)

CREATE TABLE HASA(





STREETNO CHAR(6) NOT NULL,

PRIMARY KEY (ID,PROVINCE,CITY,STREET,STREETNO),


FOREIGN KEY (PROVINCE,CITY,STREET,STREETNO) REFERENCES ADDRESS

(PROVINCE,CITY,STREET,STREETNO)

)

CREATE TABLE RONS(



PRIMARY KEY (RID,SHIFTID),

FOREIGN KEY (RID) REFERENCES RECEPTIONIST (RID),

FOREIGN KEY (SHIFTID) REFERENCES SHIFT (SHIFTID)

)

CREATE TABLE NONS(



PRIMARY KEY (NID,SHIFTID),

FOREIGN KEY (NID) REFERENCES NURSE (NID),


)


65

CREATE TABLE DONS(



PRIMARY KEY (DID,SHIFTID),

FOREIGN KEY (DID) REFERENCES DOCTOR (DID),


)

CREATE TABLE MED(

PX INTEGER NOT NULL,



MED CHAR(30) NOT NULL,

DOSAGE INTEGER NOT NULL,

MEDFROM DATE NOT NULL,

MEDTO DATE NOT NULL,

HOWOFTEN INTEGER NOT NULL,

PRIMARY KEY (PX),

UNIQUE (PID,MED),

FOREIGN KEY (PID) REFERENCES PATIENT (PID),

FOREIGN KEY (DID) REFERENCES DOCTOR (DID),

FOREIGN KEY (MED) REFERENCES MEDICATION (NAME)

)

CREATE TABLE MEDA(

PX INTEGER NOT NULL,



PRIMARY KEY (PX,SHIFTID,NID),

FOREIGN KEY (PX) REFERENCES MED (PX),

FOREIGN KEY (NID,SHIFTID) REFERENCES NONS (NID,SHIFTID)

)

CREATE TABLE BEDA(


BEDNO CHAR(3) NOT NULL,

FROM TIMESTAMP NOT NULL,

TO TIMESTAMP NOT NULL,

PRIMARY KEY (PID, BEDNO),


FOREIGN KEY (BEDNO) REFERENCES BED (NUMBER)

)

CREATE TABLE CASEDOC(




PRIMARY KEY (PID,SHIFTID,DID),


FOREIGN KEY (DID,SHIFTID) REFERENCES DONS (DID,SHIFTID)

)

CREATE TABLE SUPBY(

BEDNO CHAR(3) NOT NULL,



PRIMARY KEY (BEDNO,SHIFTID,NID),

FOREIGN KEY (BEDNO) REFERENCES BED (NUMBER),

FOREIGN KEY (NID,SHIFTID) REFERENCES NONS (NID,SHIFTID)


66

)

CREATE TABLE ADM(




ADMISSION TIMESTAMP,

PRIMARY KEY (PID),


FOREIGN KEY (RID,SHIFTID) REFERENCES RONS (RID,SHIFTID)

)

CREATE TABLE TRIAGEBY(



PRIMARY KEY (PID),


FOREIGN KEY (DID) REFERENCES DOCTOR (DID)

)

8.5 Emergency Room Interactive Queries

Query 1: The query returns an empty set if con1 is satisfied.

((SELECT RID FROM RECEPTIONIST) INTERSECT (SELECT NID FROM NURSE))

UNION

((SELECT RID FROM RECEPTIONIST) INTERSECT (SELECT DID FROM DOCTOR))

UNION

((SELECT NID FROM NURSE) INTERSECT (SELECT DID FROM DOCTOR)


SELECT EADDRESS FROM EMAIL

EXCEPT

SELECT EADDRESS FROM HASE


SELECT AREACODE,NUMBER FROM PHONENO

EXCEPT

SELECT AREACODE,NUMBER FROM HASP


SELECT PROVINCE,CITY,STREET,STRETNO FROM ADDRESS

EXCEPT

SELECT PROVINCE,CITY,STREET,STREETNO FROM HASA


SELECT ID FROM PERSON

EXCEPT


67

SELECT ID FROM HASA


((SELECT SHIFTID FROM RONS)

EXCEPT

((SELECT T.SHIFTID FROM RONS AS T,RONS AS R

WHERE T.RID <> R.RID AND T.SHIFTID = R.SHIFTID)

EXCEPT

(SELECT T.SHIFTID FROM RONS AS T,RONS AS R,RONS AS Q

WHERE T.SHIFTID = R.SHIFTID AND T.SHIFTID = Q.SHIFTID

AND T.RID <> R.RID AND T.RID<>Q.RID AND R.RID<>Q.RID))

)


(SELECT SHIFTID FROM SHIFT) EXCEPT (SELECT SHIFTID FROM RONS)

UNION

(SELECT RID FROM RECEPTIONIST) EXCEPT (SELECT RID FROM RONS)


SELECT SHIFTID FROM SHIFT

EXCEPT

SELECT T.SHIFTID FROM NONS AS T,NONS AS R

WHERE T.NID <> R.NID AND T.SHIFTID = R.SHIFTID


(SELECT SHIFTID FROM SHIFT) EXCEPT (SELECT SHIFTID FROM NONS)

UNION

(SELECT NID FROM NURSE) EXCEPT (SELECT NID FROM NONS)


((SELECT SHIFTID FROM SHIFT)

EXCEPT

(SELECT T.SHIFTID

FROM DONS AS T,DONS AS R

WHERE T.DID <> R.DID AND T.SHIFTID = R.SHIFTID))


((SELECT SHIFTID FROM SHIFT) EXCEPT (SELECT SHIFTID FROM DONS))

UNION

((SELECT DID FROM DOCTOR) EXCEPT (SELECT DID FROM DONS))


((SELECT PID FROM CASEDOC) EXCEPT (SELECT PID FROM BEDA))


68


((SELECT PID FROM BEDA) EXCEPT (SELECT PID FROM CASEDOC))


((SELECT BED.NUMBER,SHIFTID FROM BED,SHIFT)

EXCEPT

(SELECT BED.NUMBER,SHIFTID FROM BED,SUPBY

WHERE BED.NUMBER=SUPBY.BEDNO))


(SELECT FROM,TO,ADMISSION FROM SHIFT,ADM

WHERE SHIFT.SHIFTID=ADM.SHIFTID

AND (ADMISSION < FROM OR TO < ADMISSION))


((SELECT PID FROM TRIAGEBY)

EXCEPT

(SELECT TRIAGEBY.PID FROM DONS,ADM,TRIAGEBY

WHERE TRIAGEBY.PID=ADM.PID AND ADM.SHIFTID=DONS.SHIFTID

AND DONS.DID=TRIAGEBY.DID))


69

Chapter 9: Property Rental

9.1 Property Rental Informal Description

Our company arranges rentals of properties owned by both private and business owners.

We assign every property owner a unique owner number for identification, we record its

address (consisting of a street, street number, town or city, and province), the owner’s

name (consisting of first, middle, and last name for a person or name of a business), and

the owners email addresses and the owner phone numbers. For a business owner, we

record the type (description) of its business. Each property is identified by a unique

property number, we record its address and its type. Each property may be placed in

several advertisements. Each such advertisement may be displayed in many newspapers

on several dates. The newspapers are identified by unique names.

The term renter refers to a private person or a business who signed a rental

agreement for a property. Each such rental agreement is identified in our database by a

unique rental number. We record the date of the singing of the rental agreement, the

starting and ending date of the rental agreement. A renter can rent many properties. A

renter, prior to accepting the rental agreement may view the property repeatedly and we

record the date of viewing. For each renter, we record its address, its name, its email

address and phone numbers. Each renter has a unique renter number in our database.

Our agency is organized into branches and every staff member is allocated to

exactly one branch. Each branch has one manager who is a member of the staff. In our

database, we identify the staff by a unique staff number. For each staff member we record

address, name, email address, phone numbers, sex, position, and salary. Each property is

in care of one of our branches. Each renter refers to the branch that is in care of the

property it rents. Each property is overseen by a unique staff member. Each branch has an

address, phone number, and a unique branch number.


70

9.2 Property Rental Logical Model

Figure 14: The Property Rental Logical Model


71

9.3 Property Rental Physical DB2 Model

Figure 15: The Property Rental Physical Model


72

9.4 Property Rental DB2 Schema CREATE TABLE BRANCH (

BRANCH_NO CHAR(4) NOT NULL,

STREET_NO CHAR(4) NOT NULL,




POSTAL_CODE CHAR(6) NOT NULL,

MANAGER CHAR(4) NOT NULL,

PRIMARY KEY (BRANCH_NO),

CHECK (PROVINCE IN ('AL','BC','MA','NB','NF','NT','NS', 'NU',

'ON','PE','QB','SA','YU')),

CHECK (('A'<=SUBSTR(POSTAL_CODE,1,1) AND

SUBSTR(POSTAL_CODE,1,1)<='Z') AND

('0'<=SUBSTR(POSTAL_CODE,2,1) AND SUBSTR(POSTAL_CODE,2,1)<='9')

AND

('A'<=SUBSTR(POSTAL_CODE,3,1) AND SUBSTR(POSTAL_CODE,3,1)<='Z')

AND

('0'<=SUBSTR(POSTAL_CODE,4,1) AND SUBSTR(POSTAL_CODE,4,1)<='9')

AND

('A'<=SUBSTR(POSTAL_CODE,5,1) AND SUBSTR(POSTAL_CODE,5,1)<='Z')

AND

('0'<=SUBSTR(POSTAL_CODE,6,1) AND

SUBSTR(POSTAL_CODE,6,1)<='9')),

UNIQUE(MANAGER)

)

CREATE TABLE STAFF (

STAFF_NO CHAR(4) NOT NULL,

LAST_NAME CHAR(20) NOT NULL,

FIRST_NAME CHAR(10) NOT NULL,

MIDDLE_NAME CHAR(10),






SEX CHAR(1) NOT NULL,

SALARY DECIMAL(9,2) NOT NULL,

ALLOCATED_TO CHAR(4) NOT NULL,

PRIMARY KEY (STAFF_NO),

FOREIGN KEY (ALLOCATED_TO) REFERENCES BRANCH,

CHECK (PROVINCE IN ('AL','BC','MA','NB','NF','NT','NS',

'NU','ON','PE','QB','SA','YU')),

CHECK (SEX IN ('F','M','N')),

CHECK (SALARY > 0),

CHECK (('A'<=SUBSTR(POSTAL_CODE,1,1)

AND SUBSTR(POSTAL_CODE,1,1)<='Z')

AND ('0'<=SUBSTR(POSTAL_CODE,2,1)

AND SUBSTR(POSTAL_CODE,2,1)<='9')

AND ('A'<=SUBSTR(POSTAL_CODE,3,1)



73


AND SUBSTR(POSTAL_CODE,4,1)<='9')

AND ('A'<=SUBSTR(POSTAL_CODE,5,1)



AND SUBSTR(POSTAL_CODE,6,1)<='9'))

)

CREATE TABLE OWNER (

OWNER_NO CHAR(4) NOT NULL,


FIRST_NAME CHAR(10),







TYPE_OF_BUSINESS CHAR(2),

PRIMARY KEY (OWNER_NO),

CHECK (PROVINCE IN ('AL','BC','MA','NB','NF','NT','NS','NU',

'ON','PE','QB','SA','YU')),

CHECK (('A'<=SUBSTR(POSTAL_CODE,1,1) AND SUBSTR(POSTAL_CODE,1,1)<='Z')

AND ('0'<=SUBSTR(POSTAL_CODE,2,1) AND SUBSTR(POSTAL_CODE,2,1)<='9')

AND ('A'<=SUBSTR(POSTAL_CODE,3,1) AND SUBSTR(POSTAL_CODE,3,1)<='Z')



AND ('0'<=SUBSTR(POSTAL_CODE,6,1) AND SUBSTR(POSTAL_CODE,6,1)<='9')),

CHECK(TYPE_OF_BUSINESS IS NULL OR (FIRST_NAME IS NULL AND MIDDLE_NAME

IS NULL))

)

CREATE TABLE RENTER (

RENTER_NO CHAR(4) NOT NULL,


FIRST_NAME CHAR(10),







TYPE_OF_BUSINESS CHAR(2),

PRIMARY KEY (RENTER_NO),

CHECK (PROVINCE IN ('AL','BC','MA','NB','NF','NT','NS','NU','ON',

'PE','QB','SA','YU')),






AND ('0'<=SUBSTR(POSTAL_CODE,6,1) AND SUBSTR(POSTAL_CODE,6,1)<='9')),

CHECK(TYPE_OF_BUSINESS IS NULL OR (FIRST_NAME IS NULL AND MIDDLE_NAME

IS NULL))


74

)

CREATE TABLE PROPERTY (

PROPERTY_NO CHAR(4) NOT NULL,






OVERSEEN_BY CHAR(4) NOT NULL,

OWNED_BY CHAR(4) NOT NULL,

TYPE CHAR(2) NOT NULL,

PRIMARY KEY (PROPERTY_NO),

FOREIGN KEY (OVERSEEN_BY) REFERENCES STAFF,

FOREIGN KEY (OWNED_BY) REFERENCES OWNER,

CHECK (PROVINCE IN ('AL','BC','MA','NB','NF','NT','NS','NU','ON',

'PE','QB','SA','YU')),






AND ('0'<=SUBSTR(POSTAL_CODE,6,1) AND SUBSTR(POSTAL_CODE,6,1)<='9'))

)

CREATE TABLE RENTAL_AGREEMENT (


RENTAL_NO CHAR(4) NOT NULL,

SIGNING_DATE DATE NOT NULL,

STARTING_DATE DATE NOT NULL,

ENDING_DATE DATE NOT NULL,


PRIMARY KEY (PROPERTY_NO,RENTAL_NO),

FOREIGN KEY (PROPERTY_NO) REFERENCES PROPERTY,

FOREIGN KEY (RENTER_NO) REFERENCES RENTER,

CHECK (SIGNING_DATE <= STARTING_DATE),

CHECK (STARTING_DATE <= ENDING_DATE)

)

CREATE TABLE RENTER_EMAIL (

EMAIL_ADDR CHAR(20) NOT NULL,


PRIMARY KEY (EMAIL_ADDR,RENTER_NO),

FOREIGN KEY (RENTER_NO) REFERENCES RENTER

)

CREATE TABLE STAFF_EMAIL (



PRIMARY KEY (EMAIL_ADDR,STAFF_NO),

FOREIGN KEY (STAFF_NO) REFERENCES STAFF

)

CREATE TABLE OWNER_EMAIL (



PRIMARY KEY (EMAIL_ADDR,OWNER_NO),

FOREIGN KEY (OWNER_NO) REFERENCES OWNER

)


75

CREATE TABLE BRANCH_EMAIL (



PRIMARY KEY (EMAIL_ADDR,BRANCH_NO),

FOREIGN KEY (BRANCH_NO) REFERENCES BRANCH

)

CREATE TABLE RENTER_PHONE (

AREA_CODE CHAR(3) NOT NULL,

PHONE_NO CHAR(7) NOT NULL,

EXTENSION VARCHAR(5),


PRIMARY KEY (AREA_CODE,PHONE_NO,RENTER_NO),

FOREIGN KEY (RENTER_NO) REFERENCES RENTER,

CHECK(('0'<=SUBSTR(AREA_CODE,1,1) AND SUBSTR(AREA_CODE,1,1)<='9')

AND ('0'<=SUBSTR(AREA_CODE,2,1) AND SUBSTR(AREA_CODE,2,1)<='9')

AND ('0'<=SUBSTR(AREA_CODE,3,1) AND SUBSTR(AREA_CODE,3,1)<='9')),

CHECK(('0'<=SUBSTR(PHONE_NO,1,1) AND SUBSTR(PHONE_NO,1,1)<='9')

AND ('0'<=SUBSTR(PHONE_NO,2,1) AND SUBSTR(PHONE_NO,2,1)<='9')





AND ('0'<=SUBSTR(PHONE_NO,7,1) AND SUBSTR(PHONE_NO,7,1)<='9'))

)

CREATE TABLE STAFF_PHONE (





PRIMARY KEY (AREA_CODE,PHONE_NO,STAFF_NO),

FOREIGN KEY (STAFF_NO) REFERENCES STAFF,











)

CREATE TABLE OWNER_PHONE (





PRIMARY KEY (AREA_CODE,PHONE_NO,OWNER_NO),

FOREIGN KEY (OWNER_NO) REFERENCES OWNER,







76






)

CREATE TABLE BRANCH_PHONE (





PRIMARY KEY (AREA_CODE,PHONE_NO,BRANCH_NO),

FOREIGN KEY (BRANCH_NO) REFERENCES BRANCH,











)

CREATE TABLE VIEWING (



VIEWING_DATE DATE NOT NULL,

PRIMARY KEY (PROPERTY_NO,VIEWING_DATE,RENTER_NO),

FOREIGN KEY (PROPERTY_NO) REFERENCES PROPERTY,

FOREIGN KEY (RENTER_NO) REFERENCES RENTER

)

CREATE TABLE NEWSPAPER (

PAPER_NAME CHAR(20) NOT NULL,

PRIMARY KEY (PAPER_NAME)

)

CREATE TABLE ADVERTISEMENT (

PAPER_NAME CHAR(20) NOT NULL,

AD_NO CHAR(4) NOT NULL,

AD_DATE DATE NOT NULL,


PRIMARY KEY (PAPER_NAME,AD_NO),

FOREIGN KEY (PAPER_NAME) REFERENCES NEWSPAPER,

FOREIGN KEY (PROPERTY_NO) REFERENCES PROPERTY

)

ALTER TABLE BRANCH

ADD CONSTRAINT MANAGER_CNST FOREIGN KEY (MANAGER) REFERENCES

STAFF(STAFF_NO)


77

9.5 Property Rental Interactive Queries

Query 1: Give the staff number of each of the staff members whose salary is greater than

5000. Please, sort them by the staff number.

SELECT STAFF_NO FROM STAFF

WHERE SALARY > 5000

ORDER BY STAFF_NO

Query 2: Give the renter number of each of the renters who has a viewing record. Please

avoid duplications.

SELECT DISTINCT RENTER_NO FROM VIEWING

Query 3: Give the dates of all the advertisements posted in THE GLOBE AND MAIL in

2005. Please, avoid duplications.

SELECT DISTINCT AD_DATE FROM ADVERTISEMENT

WHERE PAPER_NAME = 'THE GLOBE AND MAIL' AND AD_DATE>='2005-01-01'

AND AD_DATE<='2005-12-31'

ORDER BY AD_DATE

Query 4: Give the email addresses and the renter number for all the private renters.

Please, sort them by the renter number.

SELECT EMAIL_ADDR, RENTER.RENTER_NO

FROM RENTER_EMAIL, RENTER

WHERE RENTER_EMAIL.RENTER_NO = RENTER.RENTER_NO

AND TYPE_OF_BUSINESS IS NULL

ORDER BY RENTER.RENTER_NO

Query 5: Find the properties that are already advertised but not yet rented. Please, avoid

duplications.

SELECT DISTINCT PROPERTY_NO FROM ADVERTISEMENT

WHERE PROPERTY_NO NOT IN (SELECT DISTINCT PROPERTY_NO

FROM RENTAL_AGREEMENT)

Query 6: Give the names and the branch numbers of all the staff members working in the

branch which is located in Hamilton. The names should be listed in an

alphabetic order (by last, then by first, then by middle names).

SELECT FIRST_NAME,MIDDLE_NAME,LAST_NAME,BRANCH_NO

FROM STAFF,BRANCH

WHERE STAFF.ALLOCATED_TO=BRANCH.BRANCH_NO

AND BRANCH.CITY='HAMILTON'

ORDER BY LAST_NAME,FIRST_NAME,MIDDLE_NAME


78

Query 7: Give the staff numbers and the names of all the workers who live on the same

street, city, and province as their manager. The names should be listed in an

alphabetic order (by last, then by first, then by middle names).

SELECT STAFF.STAFF_NO, FIRST_NAME, MIDDLE_NAME, LAST_NAME

FROM STAFF,(SELECT STAFF_NO, BRANCH_NO, STAFF.STREET, STAFF.CITY,

STAFF.PROVINCE FROM STAFF,BRANCH

WHERE STAFF.STAFF_NO = BRANCH.MANAGER) AS T

WHERE STAFF.ALLOCATED_TO = T.BRANCH_NO AND

STAFF.STAFF_NO != T.STAFF_NO AND

STAFF.STREET = T.STREET AND

STAFF.CITY = T.CITY AND

STAFF.PROVINCE = T.PROVINCE

ORDER BY LAST_NAME, FIRST_NAME, MIDDLE_NAME

Query 8: Find the branch number and the average salary of the branch that has the highest

average salary. Please, call the branch number as branch_no and the

average salary as avg_salary.

(SELECT ALLOCATED_TO AS BRANCH_NO, AVG(SALARY) AS AVG_SALARY

FROM STAFF GROUP BY ALLOCATED_TO)

EXCEPT

(SELECT T1.ALLOCATED_TO AS BRANCH_NO, T1.AVG_SALARY

FROM (SELECT ALLOCATED_TO, AVG(SALARY) AS AVG_SALARY

FROM STAFF GROUP BY ALLOCATED_TO) AS T1,

(SELECT ALLOCATED_TO, AVG(SALARY) AS AVG_SALARY

FROM STAFF GROUP BY ALLOCATED_TO) AS T2

WHERE T1.AVG_SALARY < T2.AVG_SALARY)

Query 9: Find the owners and renters who have 2 or more phone numbers. Call the

owner/renter number as customer_no, set the value of

type_of_customer to 'owner' if the customer is an owner, and to

'renter' if he/she is a renter. Please, only list the customer_no and

type_of_customer.

(SELECT OWNER_NO AS CUSTOMER_NO, 'OWNER' AS TYPE_OF_CUSTOMER

FROM OWNER

WHERE OWNER_NO IN (SELECT OWNER_NO FROM OWNER_PHONE

GROUP BY OWNER_NO HAVING COUNT(*) >= 2))

UNION

(SELECT RENTER_NO AS CUSTOMER_NO, 'RENTER' AS TYPE_OF_CUSTOMER

FROM RENTER

WHERE RENTER_NO IN (SELECT RENTER_NO FROM RENTER_PHONE

GROUP BY RENTER_NO HAVING COUNT(*) >= 2))

Query 10: Assuming that each advertisement costs 100 dollars, give the branch

number and the amount spent on the advertisements for each branch. Name

the branch number as branch_no, and the amount as ad_cost.

SELECT T2.ALLOCATED_TO AS BRANCH_NO, SUM(C)*100 AS AD_COST


79

FROM (SELECT PROPERTY_NO, COUNT(*) AS C

FROM ADVERTISEMENT GROUP BY PROPERTY_NO) AS T1,

(SELECT PROPERTY_NO, ALLOCATED_TO FROM PROPERTY,STAFF

WHERE PROPERTY.OVERSEEN_BY = STAFF.STAFF_NO) AS T2

WHERE T1.PROPERTY_NO = T2.PROPERTY_NO

GROUP BY T2.ALLOCATED_TO


80

Chapter 10: Software Project

10.1 Software Project Informal Description

Our website manages software projects for downloads to users. Each software project has

a unique project id (8 characters long), can be assigned one or more categories (the

categories are A, B ,C and D), has a status (D or P), and has a description (text of at most

256 characters). Some projects may depend on other projects and we keep track of the

dependency. Each project is developed and owned by a single developer (who is our

subscriber), and uploaded to our website in one or more transactions.

Our users are identified by name (at most 20 characters), email (at most 20

characters), and a unique user id (8 characters long). They can be either guest users or

subscribed users (subscribers for short). The subscribers have passwords (at most 8

characters) and we keep the date of the subscription. They need the password to access

our website to file bug reports or upload software projects or update patches. A user can

download any project, the number of downloads per user per project is recorded. The

subscribers can file bug reports for any project. Every bug identified has an id (a positive

integer) and a description (text of at most 256 characters). The bug id’s must be unique

for all bugs concerning the same project. The date of filing of a bug report is recorded.

Each bug report deals with a single project and can report a single bug. Each bug report is

made by a single subscriber.

Some of our subscribers are developers. They develop the software projects and

also software updates for their own projects. Each update for a project has an id (8

characters long), a name (at most 20 characters), a status (P or U), a description (text of at

most 256 characters), and is assigned a particular type (the type are 1, 2 and 3). Each

update for a project is created by a single developer, the one who originally created the

project. Each update patch is uploaded to our website in a transaction.

Each transaction has an id (6 characters long) and a date when it took place. The

transaction id’s must be unique for all transactions concerning the same project.


81

10.2 Software Project Logical Model

Figure 16: The Software Project Logical Model


82

10.3 Software Project Physical DB2 Model

Figure 17: The Software Project Physical Model


83

10.4 Software Project DB2 Schema CREATE TABLE USER (

USER_ID CHAR(8) NOT NULL,


EMAIL CHAR(20),

NOF_DOWNLOADS INTEGER NOT NULL,

PRIMARY KEY (USER_ID)

)

CREATE TABLE GUEST (


PRIMARY KEY (USER_ID),

FOREIGN KEY (USER_ID) REFERENCES USER

)

CREATE TABLE SUBSCRIBER (


SUBSCRDATE DATE NOT NULL,

PASSWORD CHAR(8) NOT NULL,



)

CREATE TABLE NONDEVELOPER (



FOREIGN KEY (USER_ID) REFERENCES SUBSCRIBER

)

CREATE TABLE DEVELOPER (



FOREIGN KEY (USER_ID) REFERENCES SUBSCRIBER

)

CREATE TABLE PROJECT (

PROJECT_ID CHAR(8) NOT NULL,

STATUS CHAR(1) NOT NULL,

DESCRIPTION CHAR(254),

OWNED_BY CHAR(8) NOT NULL,

PRIMARY KEY (PROJECT_ID),

FOREIGN KEY (OWNED_BY) REFERENCES DEVELOPER

)

CREATE TABLE DEPENDS (


DEPENDS_ON CHAR(8) NOT NULL,

PRIMARY KEY (PROJECT_ID,DEPENDS_ON),

FOREIGN KEY (PROJECT_ID) REFERENCES PROJECT,

FOREIGN KEY (DEPENDS_ON) REFERENCES PROJECT

)

CREATE TABLE CATEGORY (


CAT CHAR(1) NOT NULL,

PRIMARY KEY (PROJECT_ID,CAT),

FOREIGN KEY (PROJECT_ID) REFERENCES PROJECT

)

CREATE TABLE DOWNLOAD (


84



PRIMARY KEY (USER_ID,PROJECT_ID),

FOREIGN KEY (USER_ID) REFERENCES USER,


)

CREATE TABLE BUG (


BUG_ID INTEGER NOT NULL,


DATE DATE NOT NULL,

PRIMARY KEY (PROJECT_ID,BUG_ID),


)

CREATE TABLE BUGREPORT (



BUG_ID INTEGER NOT NULL,

PRIMARY KEY (PROJECT_ID,BUG_ID),

FOREIGN KEY (PROJECT_ID,BUG_ID) REFERENCES BUG,


)

CREATE TABLE TRANSACTION (

TRANSACT_ID CHAR(6) NOT NULL,


DATE DATE NOT NULL,

PRIMARY KEY (PROJECT_ID,TRANSACT_ID),


)

CREATE TABLE DEVELTRAN (




FOREIGN KEY (PROJECT_ID,TRANSACT_ID) REFERENCES TRANSACTION

)

CREATE TABLE UPDATETRAN (





)

CREATE TABLE PATCH (



PATCH_ID CHAR(8) NOT NULL,


STATUS CHAR(1) NOT NULL,


TYPE CHAR(1) NOT NULL,

PRIMARY KEY (PROJECT_ID,TRANSACT_ID,PATCH_ID),


)


85

10.5 Software Project Interactive Queries Query 1: Give user id and name of all developers who own a project.

SELECT DEVELOPER.USER_ID,USER.NAME

FROM DEVELOPER,USER

WHERE DEVELOPER.USER_ID=USER.USEr_ID

AND DEVELOPER.USER_ID IN (SELECT OWNED_BY FROM PROJECT

WHERE OWNED_BY IS NOT NULL)

Query 2: Give project id of all projects that have more than two update patches.

SELECT PROJECT_ID

FROM (SELECT PROJECT.PROJECT_ID,C

FROM PROJECT, (SELECT PROJECT_ID,COUNT(*) AS C

FROM PATCH GROUP BY PROJECT_ID) AS T

WHERE PROJECT.PROJECT_ID=T.PROJECT_ID)

WHERE C > 2

Query 3: For each project, give the number of all update patches and the number of all

downloads.

SELECT T.PROJECT_ID,PATCH_COUNT,DOWNLOAD_COUNT

FROM (SELECT PROJECT_ID,COUNT(*) AS PATCH_COUNT

FROM PATCH GROUP BY PROJECT_ID) AS T,

(SELECT PROJECT_ID,COUNT(*) AS DOWNLOAD_COUNT

FROM DOWNLOAD GROUP BY PROJECT_ID) AS S

WHERE T.PROJECT_ID=S.PROJECT_ID

Query 4: Give the project id of the projects with the most downloads.

SELECT PROJECT_ID

FROM (SELECT PROJECT_ID,COUNT(*) AS DOWNLOAD_COUNT

FROM DOWNLOAD GROUP BY PROJECT_ID),

(SELECT MAX(DC) AS MAXDC

FROM (SELECT COUNT(*) AS DC FROM DOWNLOAD

GROUP BY PROJECT_ID))

WHERE DOWNLOAD_COUNT=MAXDC

Query 5: Give the project id of the projects with the most update patches.

SELECT PROJECT_ID

FROM (SELECT PROJECT_ID,COUNT(*) AS PATCH_COUNT

FROM PATCH GROUP BY PROJECT_ID),

(SELECT MAX(PC) AS MAXPC

FROM (SELECT COUNT(*) AS PC

FROM PATCH GROUP BY PROJECT_ID))

WHERE PATCH_COUNT=MAXPC

Query 6: For each project, give project id of all projects that the project depends on.


86

SELECT *

FROM DEPENDS

Query 7: Give project id of all projects that do not depend on any other project.

(SELECT distinct PROJECT_ID FROM PROJECT)

EXCEPT

(SELECT PROJECT_ID FROM DEPENDS)

Query 8: Give the project id of the projects that depend on the most other projects.

SELECT T.PROJECT_ID

FROM (SELECT PROJECT_ID,COUNT(*) AS DEPEND_COUNT

FROM DEPENDS GROUP BY PROJECT_ID) AS T,

(SELECT MAX(DC) AS MAXDC

FROM (SELECT COUNT(*) AS DC

FROM DEPENDS GROUP BY PROJECT_ID))

WHERE T.DEPEND_COUNT = MAXDC

Query 9: Give description, bug id, and project id of all bug reports for all projects that

have most bug reports.

SELECT BUG_ID,BUG.PROJECT_ID,DESCRIPTION

FROM BUG, (SELECT T.PROJECT_ID,T.BC

FROM (SELECT PROJECT_ID,COUNT(*) AS BC

FROM BUG GROUP BY PROJECT_ID) AS T,

(SELECT MAX(BC1) AS MAXBC

FROM (SELECT COUNT(*) AS BC1

FROM BUG GROUP BY PROJECT_ID))

WHERE T.BC=MAXBC) AS S

WHERE BUG.PROJECT_ID=S.PROJECT_ID

Query 10: Give user id of a developer with the least amount of bug reports.

SELECT USER_ID

FROM (SELECT USER_ID,COUNT(*) AS BPD

FROM (SELECT BUG_ID,BUG.PROJECT_ID,USER_ID

FROM BUG,PROJECT,DEVELOPER

WHERE BUG.PROJECT_ID=PROJECT.PROJECT_ID

AND OWNED_BY=USER_ID)

GROUP BY USER_ID)

WHERE BPD IN (SELECT MIN(BPD) AS MINBPD

FROM (SELECT USER_ID,COUNT(*) AS BPD

FROM (SELECT BUG_ID,BUG.PROJECT_ID,USER_ID

FROM BUG,PROJECT,DEVELOPER

WHERE BUG.PROJECT_ID=PROJECT.PROJECT_ID


87

Chapter 11:Tour Operator System

10.1 Tour Operator System Informal Description

The system need to keep track of people. For each person, it records all his/her address, of

which exactly one is designated as the mailing address (so each person has at least one

address). Each address consists of country, province/state, city, street, street number, P.O.

Box number, and a list (possible empty) of phone numbers to the location of the address

and a list (possible empty) of fax numbers to the location of the address. In addition to the

list of addresses for each person it records a list (possible empty) of cell phone numbers

and a list (possible empty) of email address. Each person in the database can be an old

customer (have taken a tour of the company), a current customer (is booked to take a tour

or is on a tour right now), a tour guide, an employee (works for the tour company), or any

mixture of these (for instance an employee can take a tour and so can be a customer as

well, or an employee can work as a tour guide for a particular tour and hence be an

employee and a guide at the same time etc.). The sex and age of each person must also be

recorded, a date-of-birth is optional for an external worker, a contract reference for each

of the tours the guide is doing must be included. A guide contract references the tour (see

below) and the total amount the tour guide will be paid for the tour. The guides do not

pay for the accommodation and the meals.

The system also keeps track of all tours, past and future. Each tour has a unique

designation, itinerary, guide (at least one, but may be more than one), its status

(completed, in-progress, in-the-future), and the list of participants (not including the

guides). The itinerary consists of list of the dates the tour covers and for each date it

includes the place of breakfast, the place of lunch, the place of diner, and the place of

accommodation. For each of the places there is a contract reference. Each day in the

itinerary also includes and a simple English description of the activities during that day.

An accommodation can be a hotel, or a rented room or rooms from a rental

company, or a rented room or rooms from a private person. A meal (breakfast, lunch,

dinner) can be in hotel, restaurant, or a private place. The contract for accommodation or

meal must bear the date of the contract becomes valid, the date or dates it covers, what

the contract is for (accommodation, breakfast, lunch, dinner) if the pricing is per person

or per group or per room or per the whole facility, per night or per a certain period and the

corresponding price. It also may stipulate the minimum and the maximum of people for

the accommodation/meal for each day it covers, financial penalty if less than minimum


88

uses the accommodation. All prices are assumed to be in Canadian dollars, not conversion

is involved, regardless where the place is. Each place is identified by a single address.

Each provider of accommodation or meal has a unique designation.


89

11.2 Tour Operator System Logical Model

Figure 18: The Tour Operator System Logical Model


90

11.3 Tour Operator System Physical DB2 Model

Figure 19: The Tour Operator System Physical Model


91

11.4 Tour Operator System DB2 schema

CREATE TABLE PERSONS(

PID SMALLINT NOT NULL,

NAME VARCHAR(20) NOT NULL,

SEX CHAR(1) NOT NULL CHECK (SEX IN ('M','F')),

AGE SMALLINT NOT NULL CHECK (AGE BETWEEN 0 AND 150),

BIRTHDATE DATE NOT NULL,

PRIMARY KEY(PID),

CONSTRAINT AGECHECK CHECK (2003-YEAR(BIRTHDATE)=AGE)

)

CREATE TABLE HASE(


EADDR VARCHAR(30) NOT NULL,

PRIMARY KEY(PID,EADDR),

FOREIGN KEY (PID) REFERENCES PERSONS (PID)

)

CREATE TABLE HASCE(


PHONENO VARCHAR(20) NOT NULL,

PRIMARY KEY(PID,PHONENO),


)

CREATE TABLE ADDRESSES(

adID SMALLINT NOT NULL,

COUNTRY CHAR(3) NOT NULL,


STREET VARCHAR(10),

STRNO VARCHAR(10),

POBox VARCHAR(10),

PRIMARY KEY(adID),

CONSTRAINT ADDRCHECK1

CHECK (NOT ((STREET IS NOT NULL) AND (POBox IS NOT NULL))),


CHECK (NOT ((STRNO IS NOT NULL) AND (POBox IS NOT NULL))),


CHECK (NOT ((STREET IS NOT NULL) AND (STRNO IS NULL))),


CHECK (NOT ((STRNO IS NOT NULL) AND (STREET IS NULL)))

)

CREATE TABLE HASA(



PRIMARY KEY(PID,adID),

FOREIGN KEY (PID) REFERENCES PERSONS (PID),

FOREIGN KEY (adID) REFERENCES ADDRESSES (adID)

)

CREATE TABLE PHONES(


PRIMARY KEY(PHONENO)

)

CREATE TABLE FAXES(



92

PRIMARY KEY(PHONENO)

)

CREATE TABLE HASP(



PRIMARY KEY(PHONENO),

FOREIGN KEY (adID) REFERENCES ADDRESSES (adID),

FOREIGN KEY (PHONENO) REFERENCES PHONES (PHONENO)

)

CREATE TABLE HASF(



PRIMARY KEY(PHONENO),


FOREIGN KEY (PHONENO) REFERENCES FAXES (PHONENO)

)

CREATE TABLE EMPLOYEES(


PRIMARY KEY(PID),


)

CREATE TABLE CUSTOMERS(


PRIMARY KEY(PID),


)

CREATE TABLE GUIDES(


PRIMARY KEY(PID),


)

CREATE TABLE TOURS(

DESIG VARCHAR(5) NOT NULL,

STATUS CHAR(3) NOT NULL CHECK (STATUS IN ('P','I','F')),

PRIMARY KEY(DESIG)

)

CREATE TABLE PARTICIP(



PRIMARY KEY (PID,DESIG),

FOREIGN KEY (PID) REFERENCES CUSTOMERS (PID),

FOREIGN KEY (DESIG) REFERENCES TOURS (DESIG)

)

CREATE TABLE HASCO(




PRIMARY KEY (PID,DESIG),

FOREIGN KEY (PID) REFERENCES GUIDES (PID),


)

CREATE TABLE ITINERARIES(


DATE DATE NOT NULL,


93

DESCR VARCHAR(256),

PRIMARY KEY (DESIG,DATE),


)

CREATE TABLE PROVIDERS(

DESIGN VARCHAR(10) NOT NULL,

PRIMARY KEY (DESIGN)

)

CREATE TABLE PLACES(


DESIGN VARCHAR(10) NOT NULL,

PRIMARY KEY (adID),


FOREIGN KEY (DESIGN) REFERENCES PROVIDERS (DESIGN)

)

CREATE TABLE ISBP(


VALID_DATE DATE NOT NULL,

PRICING CHAR(1) NOT NULL CHECK (PRICING IN ('G','P')),

FROMD DATE NOT NULL,

TOD DATE NOT NULL,

MINP SMALLINT NOT NULL,

MAXP SMALLINT NOT NULL,

PENALTY DECIMAL(9,2) NOT NULL,


DATE DATE NOT NULL,



FOREIGN KEY (DESIG,DATE) REFERENCES ITINERARIES (DESIG,DATE),

FOREIGN KEY (adID) REFERENCES PLACES (adID),

CONSTRAINT ISBP_DATE1 CHECK (VALID_DATE < FROMD),

CONSTRAINT ISBP_DATE2 CHECK (FROMD <= TOD),

CONSTRAINT ISBP_DATE3 CHECK (FROMD <= DATE),

CONSTRAINT ISBP_DATE4 CHECK (DATE <= TOD),

CONSTRAINT ISBP_PER1 CHECK (MINP <= MAXP)

)

CREATE TABLE ISLP(





TOD DATE NOT NULL,





DATE DATE NOT NULL,





CONSTRAINT ISLP_DATE1 CHECK (VALID_DATE < FROMD),

CONSTRAINT ISLP_DATE2 CHECK (FROMD <= TOD),

CONSTRAINT ISLP_DATE3 CHECK (FROMD <= DATE),


94

CONSTRAINT ISLP_DATE4 CHECK (DATE <= TOD),

CONSTRAINT ISLP_PER1 CHECK (MINP <= MAXP)

)

CREATE TABLE ISDP(





TOD DATE NOT NULL,





DATE DATE NOT NULL,





CONSTRAINT ISDP_DATE1 CHECK (VALID_DATE < FROMD),

CONSTRAINT ISDP_DATE2 CHECK (FROMD <= TOD),

CONSTRAINT ISDP_DATE3 CHECK (FROMD <= DATE),

CONSTRAINT ISDP_DATE4 CHECK (DATE <= TOD),

CONSTRAINT ISDP_PER1 CHECK (MINP <= MAXP)

)

CREATE TABLE ISSP(





TOD DATE NOT NULL,





DATE DATE NOT NULL,



FOREIGN KEY (DESIG,DATE) REFERENCES

ITINERARIES (DESIG,DATE),


CONSTRAINT ISSP_DATE1 CHECK (VALID_DATE < FROMD),

CONSTRAINT ISSP_DATE2 CHECK (FROMD <= TOD),

CONSTRAINT ISSP_DATE3 CHECK (FROMD <= DATE),

CONSTRAINT ISSP_DATE4 CHECK (DATE <= TOD),

CONSTRAINT ISSP_PER1 CHECK (MINP <= MAXP)

)

11.5 Tour Operator System Interactive Queries

Query 1: list all customers, old and current.

SELECT CUSTOMERS.PID,NAME FROM CUSTOMERS,PERSONS


95

WHERE CUSTOMERS.PID=PERSONS.PID

Query 2: List all customers with all their addresses.

SELECT CUSTOMERS.PID,NAME,COUNTRY,CITY,STREET,STRNO,POBox

FROM CUSTOMERS,PERSONS,ADDRESSES,HASA

WHERE CUSTOMERS.PID=HASA.PID

AND HASA.adID=ADDRESSES.adID AND CUSTOMERS.PID=PERSONS.PID

Query 3: For a given guide, find all the tours he/she guided or will guide and the amount

he/she got/will get for the guiding the tour

SELECT HASCO.PID,NAME,DESIG,AMOUNT FROM HASCO,PERSONS

WHERE HASCO.PID=0 AND PERSONS.PID=0

Query 4: List all customers that are also guides.

SELECT CUSTOMERS.PID,NAME FROM PERSONS,CUSTOMERS,GUIDES

WHERE CUSTOMERS.PID=GUIDES.PID AND CUSTOMERS.PID=PERSONS.PID

Query 5: List all guides that guided a tour that had a lunch in a given place.

SELECT DISTINCT HASCO.PID,NAME FROM HASCO,ISLP,PERSONS

WHERE HASCO.DESIG=ISLP.DESIG AND ISLP.adID=100 AND HASCO.PID=PERSONS.PID

Query 6: List all contracts that cover dinners in a given place.

SELECT DISTINCT AMOUNT,VALID_DATE,PRICING,FROMD,TOD,

MINP,MAXP,PENALTY,DESIG

FROM ISDP WHERE adID=103

Query 7: List all providers that provide sleeping accommodation.

SELECT DISTINCT PLACES.DESIGN

FROM PLACES,ISSP WHERE PLACES.adID=ISSP.adID

Query 8: List the tours that will have breakfast at a given place on a given date .

SELECT TOURS.DESIG FROM TOURS,ISBP

WHERE TOURS.DESIG=ISBP.DESIG AND ISBP.DATE='12/18/2002' AND adID=100

Query 9: List all employees that have guided or will guide a tour or who have taken or

will taken a tour.

SELECT DISTINCT EMPLOYEES.PID,NAME

FROM EMPLOYEES,GUIDES,CUSTOMERS,PERSONS

WHERE (EMPLOYEES.PID=GUIDES.PID OR EMPLOYEES.PID=CUSTOMERS.PID)

AND EMPLOYEES.PID=PERSONS.PID

Query 10: List all customers booked for a tour starting later or on a given date.


96

SELECT DISTINCT PARTICIP.PID,NAME FROM PARTICIP,ITINERARIES,PERSONS

WHERE PARTICIP.DESIG=ITINERARIES.DESIG

AND ITINERARIES.DATE >= '12/18/2003' AND PARTICIP.PID=PERSONS.PID

AND NOT EXISTS (SELECT ITINERARIES.DATE FROM ITINERARIES

WHERE ITINERARIES.DATE < '12/18/2003'

AND ITINERARIES.DESIG=PARTICIP.DESIG)


97

Chapter 12: Warehouse System

12.1 Warehouse System Informal Description

Our company has several warehouses, each warehouse is designated by a unique 4-letter

symbol (by a letter we mean a..z and A..Z). Each warehouse has several bins that are

identified uniquely by numbers (unsigned integers), i.e. each warehouse has bins 0, 1, 2,

3, … Each bin has a particular capacity. In our warehouses (more precisely in the bins in

our warehouses) we store parts. Each part is designated by a unique part number (a 5-

symbol sequence of digits and letters). Several parts together can form another part. We

call such a part “assembly”. In the warehouses we store only the constituent parts, but we

record the assemblies in our database as it were a part. Assemblies cannot be parts of

other assemblies. A part can be a constituent part in at most in one assembly parts arrive

in batches. Each batch for a particular part has a unique batch number (unsigned integer)

and arrives on a particular date. Each batch has a size, i.e. the number of items in the

batch. All items from the same batch are stored together in the same bin (no batch is

stored in more than 1 bin). Each item in a batch has a unique item number (unsigned

integer). For example: part A1, batch 27, item 1 or part A1, batch 23, item 1 etc.

When a batch arrives, its date-in is recorded. A particular manager checks its

arrival, and this fact must be recorded in the database.

Some parts may be backordered. A part can be backordered only by a manager.

The manager, the date of the backorder are recorded, and also the quantity backordered.

When a backorder shipment arrives, the backorder’s remaining quantity is updated (the

number of items arrived is subtracted from the remaining quantity), and if it is less or

equal to 0, the backorder is deleted, but must be kept for record. There may be only a

single current (active) backorder for any parts. Assemblies cannot be backordered, only

their constituent parts.

When an item is shipped out of the warehouse, its date-out is recorded together

with the employee who checked its shipping.

Employee has a unique employee number (a 6-digit number), phone number(s) (it

consists of a 3-digit area code and a 6-digit number an employee can have 0 to many

phone numbers), name(s) (it consists of an up=to-10-characters fist name, an up-to-10-

characters middle name, and an up-to-20-characters last name, an employee can have 1 to

many names), address(s) (it consists of an up-to-6-characters street number, an up-to-20-


98

characters street name, an up-to-20-characters city name, and a 2-character abbreviation

of the province, an employee can have 1 to many address). Some of the employees are

managers. Every employee who is not a manager works under supervision of a single

manager. Managers do not work under other managers.


99

12.2 Warehouse System Logical Model

Figure 20: The Warehouse System Logical Model


100

12.3 Warehouse System Physical DB2 Model

Figure 21: The Warehouse System Physical Model


101

12.4 Warehouse System DB2 Schema

CREATE TABLE EMPLOYEE (

EMPLOYEENO CHAR(6) NOT NULL,

PRIMARY KEY (EMPLOYEENO),

CHECK (('0' <= SUBSTR(EMPLOYEENO,1,1) AND SUBSTR(EMPLOYEENO,1,1) <= '9')

AND ('0' <= SUBSTR(EMPLOYEENO,2,1) AND SUBSTR(EMPLOYEENO,2,1) <= '9')




AND ('0' <= SUBSTR(EMPLOYEENO,6,1) AND SUBSTR(EMPLOYEENO,6,1) <= '9'))

)

CREATE TABLE MANAGER (



FOREIGN KEY (EMPLOYEENO) REFERENCES EMPLOYEE

)

CREATE TABLE WORKER (


MANAGERNO CHAR(6) NOT NULL,


FOREIGN KEY (EMPLOYEENO) REFERENCES EMPLOYEE,

FOREIGN KEY (MANAGERNO) REFERENCES MANAGER(EMPLOYEENO)

)

CREATE TABLE HASPHONE (




PRIMARY KEY (EMPLOYEENO,AREA_CODE,NUMBER),

FOREIGN KEY (EMPLOYEENO) REFERENCES EMPLOYEE,

CHECK (('0' <= SUBSTR(AREA_CODE,1,1) AND SUBSTR(AREA_CODE,1,1) <= '9')

AND ('0' <= SUBSTR(AREA_CODE,2,1) AND SUBSTR(AREA_CODE,2,1) <= '9')

AND ('0' <= SUBSTR(AREA_CODE,3,1) AND SUBSTR(AREA_CODE,3,1) <= '9')),

CHECK (('0' <= SUBSTR(NUMBER,1,1) AND SUBSTR(NUMBER,1,1) <= '9')

AND ('0' <= SUBSTR(NUMBER,2,1) AND SUBSTR(NUMBER,2,1) <= '9')




AND ('0' <= SUBSTR(NUMBER,6,1) AND SUBSTR(NUMBER,6,1) <= '9'))

)

CREATE TABLE HASNAME (


FIRST VARCHAR(10) NOT NULL,

MIDDLE VARCHAR(10),

LAST VARCHAR(20) NOT NULL,

PRIMARY KEY (EMPLOYEENO,FIRST,LAST),


)

CREATE TABLE HASADDRESS (


STRNO VARCHAR(6) NOT NULL,

STREET VARCHAR(20) NOT NULL,


102



PRIMARY KEY (EMPLOYEENO,STRNO,STREET,CITY,PROVINCE),


)

CREATE TABLE PART (

PARTNO CHAR(5) NOT NULL,

PRIMARY KEY (PARTNO),

CHECK ((('0' <= SUBSTR(PARTNO,1,1) AND SUBSTR(PARTNO,1,1) <= '9')

OR ('a' <= SUBSTR(PARTNO,1,1) AND SUBSTR(PARTNO,1,1) <= 'z')

OR ('A' <= SUBSTR(PARTNO,1,1) AND SUBSTR(PARTNO,1,1) <= 'Z'))

AND (('0' <= SUBSTR(PARTNO,2,1) AND SUBSTR(PARTNO,2,1) <= '9')


OR ('A' <= SUBSTR(PARTNO,2,1) AND SUBSTR(PARTNO,2,1) <= 'Z'))

AND (('0' <= SUBSTR(PARTNO,3,1) AND SUBSTR(PARTNO,3,1) <= '9')


OR ('A' <= SUBSTR(PARTNO,3,1) AND SUBSTR(PARTNO,3,1) <= 'Z')))

)

CREATE TABLE SUBPART (


ASSEMBLYNO CHAR(5) NOT NULL,

PRIMARY KEY (PARTNO,ASSEMBLYNO),

FOREIGN KEY (PARTNO) REFERENCES PART,

FOREIGN KEY (ASSEMBLYNO) REFERENCES PART(PARTNO)

)

CREATE TABLE WAREHOUSE (

WAREHOUSEID CHAR(4) NOT NULL,

PRIMARY KEY (WAREHOUSEID),

CHECK ((('a' <= SUBSTR(WAREHOUSEID,1,1)

AND SUBSTR(WAREHOUSEID,1,1) <= 'z')

OR('A' <= SUBSTR(WAREHOUSEID,1,1) AND SUBSTR(WAREHOUSEID,1,1) <= 'Z'))

AND(('a' <= SUBSTR(WAREHOUSEID,2,1) AND SUBSTR(WAREHOUSEID,2,1) <= 'z')

OR ('A' <= SUBSTR(WAREHOUSEID,2,1) AND SUBSTR(WAREHOUSEID,2,1) <= 'Z'))


OR ('A' <= SUBSTR(WAREHOUSEID,3,1) AND SUBSTR(WAREHOUSEID,3,1) <= 'Z'))


OR('A' <= SUBSTR(WAREHOUSEID,4,1) AND SUBSTR(WAREHOUSEID,4,1) <= 'Z')))

)

CREATE TABLE BIN (


BINNO INTEGER NOT NULL,

CAPACITY INTEGER NOT NULL,

PRIMARY KEY (WAREHOUSEID,BINNO),

FOREIGN KEY (WAREHOUSEID) REFERENCES WAREHOUSE,

CHECK (BINNO >= 0),

CHECK (CAPACITY >= 0)

)

CREATE TABLE BATCH (


BATCHNO INTEGER NOT NULL,

SIZE INTEGER NOT NULL,

DATE_IN DATE NOT NULL,

MANAGERNO CHAR(6) NOT NULL,



103

BINNO INTEGER NOT NULL,

PRIMARY KEY (PARTNO,BATCHNO),


FOREIGN KEY (MANAGERNO) REFERENCES MANAGER(EMPLOYEENO),

FOREIGN KEY (WAREHOUSEID,BINNO) REFERENCES BIN,

CHECK (BATCHNO >= 0),

CHECK (SIZE > 0)

)

CREATE TABLE ITEM (


BATCHNO INTEGER NOT NULL,

ITEMNO INTEGER NOT NULL,

CHECKED_OUT CHAR(6),

DATE_OUT DATE,

PRIMARY KEY (PARTNO,BATCHNO,ITEMNO),

FOREIGN KEY (PARTNO,BATCHNO) REFERENCES BATCH,

FOREIGN KEY (CHECKED_OUT) REFERENCES EMPLOYEE(EMPLOYEENO),

CHECK (ITEMNO >= 0)

)

CREATE TABLE CURRENT_BACKORDER (


ORIG_QUANTITY INTEGER NOT NULL,

REMAINING_QUANTITY INTEGER NOT NULL,

BO_DATE DATE NOT NULL,

BACKORDERED_BY CHAR(6) NOT NULL,

PRIMARY KEY (PARTNO),


FOREIGN KEY (BACKORDERED_BY) REFERENCES MANAGER(EMPLOYEENO),

CHECK (ORIG_QUANTITY > 0),

CHECK (REMAINING_QUANTITY >= 0 AND REMAINING_QUANTITY <= ORIG_QUANTITY)

)

CREATE TABLE OLD_BACKORDER (


ORIG_QUANTITY INTEGER NOT NULL,

BO_DATE DATE NOT NULL,

BACKORDERED_BY CHAR(6) NOT NULL,

FULFILLED TIMESTAMP NOT NULL,

PRIMARY KEY (PARTNO,BO_DATE,FULFILLED),


FOREIGN KEY (BACKORDERED_BY) REFERENCES MANAGER(EMPLOYEENO),

CHECK (DATE(FULFILLED) >= BO_DATE)

)

12.5 Warehouse System Interactive Queries

Query 1: give all employee_no for all the workers that work under manager with the first

name Tony7 and the last name Tona7 with no middle name.

SELECT WORKER.EMPLOYEENO

FROM WORKER,

(SELECT MANAGER.EMPLOYEENO


104

FROM MANAGER,HASNAME

WHERE MANAGER.EMPLOYEENO=HASNAME.EMPLOYEENO

AND HASNAME.FIRST='TONY7' AND HASNAME.MIDDLE IS NULL

AND HASNAME.LAST='TONA7') AS T

WHERE WORKER.MANAGERNO=T.EMPLOYEENO

Query 2: give all the names and employee_no for all the workers the names should be

listed in an alphabetic order (by last, then by first, then by middle)

SELECT WORKER.EMPLOYEENO,FIRST,MIDDLE,LAST ROM WORKER,HASNAME

WHERE WORKER.EMPLOYEENO=HASNAME.EMPLOYEENO

ORDER BY LAST, FIRST, MIDDLE

Query 3: give all the phones and employee_no for all the managers.

SELECT MANAGER.EMPLOYEENO,AREA_CODE,NUMBER

FROM MANAGER,HASPHONE

WHERE MANAGER.EMPLOYEENO=HASPHONE.EMPLOYEENO

Query 4: list all parts that are assemblies they should be listed in a lexicographic order.

SELECT DISTINCT ASSEMBLYNO FROM SUBPART ORDER BY ASSEMBLYNO

Query 5: for each manager, list all current backorders done by the manager.

SELECT MANAGER.EMPLOYEENO,PARTNO,ORIG_QUANTITY,REMAINING_QUANTITY,

BO_DATE, BACKORDERED_BY

FROM MANAGER,CURRENT_BACKORDER

WHERE MANAGER.EMPLOYEENO=CURRENT_BACKORDER.BACKORDERED_BY

Query 6: For each manager, list all current and old backorders done by the manager. For

each backorder you have to list the part_no, backorder date, and fulfilled date.

For current backorders, list a phony fulfilled date '2000-01-01'.

(SELECT MANAGER.EMPLOYEENO, PARTNO, BO_DATE, '2000-01-01' AS FD

FROM MANAGER,CURRENT_BACKORDER

WHERE MANAGER.EMPLOYEENO=CURRENT_BACKORDER.BACKORDERED_BY

UNION

SELECT MANAGER.EMPLOYEENO, PARTNO, BO_DATE,DATE(FULFILLED) AS FD

FROM MANAGER,OLD_BACKORDER

WHERE MANAGER.EMPLOYEENO=OLD_BACKORDER.BACKORDERED_BY

) ORDER BY EMPLOYEENO

Query 7: For each warehouse bin, give the remaining capacity of the bin. Call the

remaining capacity remaining_capacity.

SELECT T1.WAREHOUSEID, T1.BINNO, CAPACITY-C AS REMAINING_CAPACITY

FROM (SELECT WAREHOUSEID, BINNO, CAPACITY FROM BIN) AS T1,


105

(SELECT BATCH.WAREHOUSEID, BATCH.BINNO, COUNT(ITEMNO) AS C

FROM ITEM, BATCH WHERE ITEM.PARTNO=BATCH.PARTNO

AND ITEM.BATCHNO=BATCH.BATCHNO

GROUP BY BATCH.WAREHOUSEID, BATCH.BINNO) AS T2

WHERE T1.WAREHOUSEID=T2.WAREHOUSEID AND T1.BINNO=T2.BINNO

Query 8: Give employee_no and number of workers managed for all the managers with

The smallest number of workers managed.

SELECT MANAGERNO, NUMBER_MANAGED

FROM (SELECT MANAGERNO,COUNT(*) AS NUMBER_MANAGED

FROM WORKER GROUP BY MANAGERNO) AS T1

EXCEPT

SELECT T1.MANAGERNO, T1.NUMBER_MANAGED

FROM (SELECT MANAGERNO,COUNT(*) AS NUMBER_MANAGED

FROM WORKER GROUP BY MANAGERNO) AS T1,

(SELECT MANAGERNO,COUNT(*) AS NUMBER_MANAGED

FROM WORKER GROUP BY MANAGERNO) AS T2

WHERE T1.NUMBER_MANAGED > T2.NUMBER_MANAGED


106

Conclusion and Future Work

In this thesis we presented a suite of ten projects designed for undergraduate students

taking a database course to practise certain aspects of dealing with relational databases.

Their main utility and our main objective was to provide projects that are not too simple

and that are not too complex, projects that would allow students to practise the most

important aspects of dealing with databases. To this end, each project takes a student

through four major aspects of dealing with databases:

• Design of the database.

• Implementation of the database.

• SQL querying of the installed database.

• Application programming.

Our projects address some of what we consider most important issues associated with

each of the four major aspects, and, of course, leave many other issues untouched.

• In design, every project practises the ER/relational modeling in the context of

ERwin modeling tool.

• In implementation, every project practises the creation of the DB2 schema derived

from the ER/relational model.

• In SQL querying, in every project simple to moderately complex SQL queries are

practised in interactive form.

• In application programming, C and embedded SQL are used. For pedagogical

reasons, queries that were first practiced in the interactive querying are used in the

application program to illustrate and highlight the differences and the similarities

of ESQL and SQL.

For the future work, Oracle and Microsoft SQL Server in addition to DB2 must be

included. One of the important omissions in the projects concerns normalization on the

level of ER/relational modeling and a problem related to, the intentional de-

normalization of the design for performance gains. This omission must be addressed in

the future work on this suite of projects. In order to allow more complex applications,

ESQL-based application programming in C++ should be also considered for inclusion in

near future.


107

Bibliography

[1] E.F. Codd, A Relational Model of Data for Large Shared Data Banks,

Communications of ACM, Volume 13, Number 6, 1970, 377--387

[2] Pioneer calls relational database technology obsolete,

http://www.computerworlduk.com/news/applications/5059/pioneer-calls-relational-

database-technology-obsolete/

[3] Michael Stonebraker, The End of a DBMS Era (Might be Upon Us)

http://cacm.acm.org/blogs/blog-cacm/32212-the-end-of-a-dbms-era-might-be-upon-

us/fulltext

[4] Peter Pin-shan Chen, The Entity-Relationship Model: Toward a Unified View of

Data, ACM Transactions on Database Systems, Volume 1, 1976, 9--36

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